Introduction To Three Dimensional Geometry - Class 11 Engineering Maths - Extra Questions

If a point is in the $$XZ$$ plane then its $$y$$ -coordinate is zero

If distance between given points is $$d$$ then,
$$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}=\sqrt{(-1-0)^2+(1-5)^2+(3-6)^2}=\sqrt{1+16+9}=\sqrt{26}$$

The distance between points $$P(x_{1}, y _{1}, z_{1})$$ and  $$Q(x_{2}, y _{2}, z_{2})$$   is given by
$$PQ =$$ $$\displaystyle \sqrt{\left ( x_{2} -x_{1}\right )^{2}+\left ( y_{2} -y_{1}\right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}$$

(i) Distance between points $$(2, 3, 5)$$ and $$(4, 3, 1)$$
$$=$$ $$\displaystyle \sqrt{\left ( 4-2 \right )^{2}+\left ( 3-3 \right )^{2}+\left ( 1-5 \right )^{2}}$$
$$=$$ $$\displaystyle \sqrt{\left ( 2 \right )^{2}+\left ( 0 \right )^{2}+\left ( -4 \right )^{2}}$$
$$=$$ $$\displaystyle \sqrt{4+16}$$
$$=$$ $$\displaystyle \sqrt{20}$$
$$=$$ $$\displaystyle 2\sqrt{5}$$

(ii) Distance between points $$(-3, 7, 2)$$ and $$(2, 4, -1)$$
$$=$$ $$\displaystyle \sqrt{\left ( 2+3 \right )^{2}+\left ( 4-7 \right )^{2}+\left ( -1-2 \right )^{2}}$$
$$=$$ $$\displaystyle \sqrt{\left ( 5 \right )^{2}+\left ( -3 \right )^{2}+\left ( -3 \right )^{2}}$$
$$=$$ $$\displaystyle \sqrt{25+9+9}$$
$$=$$ $$\displaystyle \sqrt{43}$$

(iii) Distance between points $$(-1, 3, -4)$$ and $$(1, -3, 4)$$
$$=$$ $$\displaystyle \sqrt{\left ( 1+1 \right )^{2}+\left ( -3-3 \right )^{2}+\left ( 4+4 \right )^{2}}$$
$$=$$ $$\displaystyle \sqrt{\left ( 2 \right )^{2}+\left ( -6 \right )^{2}+\left ( 8 \right )^{2}}$$
$$=$$ $$\displaystyle \sqrt{4+36+64}=\sqrt{104}$$
$$=2\sqrt{26}$$

(iv) Distance between points $$(2, -1, 3)$$ and $$(-2, 1, 3)$$
$$=$$ $$\displaystyle \sqrt{\left ( -2-2 \right )^{2}+\left ( 1+1 \right )^{2}+\left ( 3-3 \right )^{2}}$$
$$=$$ $$\displaystyle \sqrt{\left ( -4 \right )^{2}+\left ( 2 \right )^{2}+\left ( 0 \right )^{2}}$$
$$=$$ $$\displaystyle \sqrt{16+4}$$
$$=$$ $$\displaystyle \sqrt{20}$$
$$=$$ $$\displaystyle 2\sqrt{5}$$

The $$x$$ -coordinate $$y$$ -coordinate and $$z$$ -coordinate of point $$(1, 2, 3)$$ are all positive.
Therefore this point lies in octant I

Let $$R$$ and $$S$$ be the points of trisection of line segment $$PQ$$ .

The equation of the line joining the points $$(-1,1)$$ and $$(5,7)$$ is given by, 
$$\displaystyle y-1= \frac{7-1}{5+1}(x+1)$$
$$\displaystyle y-1= \frac{6}{6}(x+1)$$
$$\displaystyle x-y+2= 0$$ ......(1)
and the equation of the given line is
$$x+y-4\displaystyle = 0$$ .....(2)
Thus point of intersection of lines (1) and (2) is given by,
$$x\displaystyle = 1$$ and $$y\displaystyle = 3$$
Let point $$(1,3)$$ divide the line segment joining $$(-1,1)$$ and $$(5,7)$$ in the ration $$1:k$$ .
Accordingly, by section formula,
$$\displaystyle (1,3)= \left \{ \frac{k(-1)+1(5)}{1+k},\frac{k(1)+1(7)}{1+k} \right \}$$
$$\displaystyle \Rightarrow (1,3)= \left \{ \frac{-k+5}{1+k},\frac{k+7}{1+k} \right \}$$
$$\displaystyle \Rightarrow  \frac{-k+5}{1+k}= 1,\frac{k+7}{1+k}= 3$$
$$\displaystyle \Rightarrow  \frac{-k+5}{1+k}= 1$$
$$\displaystyle \Rightarrow  -k+5= 1+k$$
$$\displaystyle \Rightarrow  2k= 4 \Rightarrow  k= 2$$
Thus, the line joining the points $$(-1,1)$$ and $$(5,7)$$ is divided by line $$x+y=4$$ in the ratio $$1:2$$ .

The coordinates of the centroid of  $$\displaystyle \bigtriangleup PQR$$
$$=\displaystyle \left ( \frac{2a-4-8}{3},\frac{2+3b+14}{3},\frac{6-10+2c}{3} \right )=\left ( \frac{2a+4}{3},\frac{3b+16}{3},\frac{2c-4}{3} \right )$$
It is given that origin is the centroid of  $$\displaystyle \bigtriangleup PQR$$
$$\displaystyle \therefore$$   $$\displaystyle \left ( 0,0,0 \right )=\left ( \frac{2a+4}{3},\frac{3b+16}{3},\frac{2c-4}{3} \right )$$
$$\displaystyle \Rightarrow \frac{2a+4}{3}=0,\frac{3b+16}{3}=0   \text{and}   \frac{2c-4}{3}=0$$
$$\displaystyle \Rightarrow a=-2, b=-\frac{16}{3}$$ and $$c=2$$

Let $$A (0, b, 0)$$ be the point on the $$y$$ -axis at a distance of  $$\displaystyle 5\sqrt{2}$$ from point  $$P (3, -2, 5)$$ .
We have given, $$AP = \displaystyle 5\sqrt{2}$$
$$\displaystyle \Rightarrow  AP^{2}=50$$
$$\displaystyle \Rightarrow \left ( 3-0 \right )^{2}+\left ( -2-B \right )^{2}+\left ( 5-0 \right )^{2}=50$$
$$\displaystyle \Rightarrow 9+4+b^{2}+4b+25=50$$
$$\displaystyle \Rightarrow b^{2}+4b-12=0$$
$$\displaystyle \Rightarrow b^{2}+6b-2b-12=0$$
$$\displaystyle \Rightarrow \left ( b+6 \right )\left ( b-2 \right )=0$$
$$\displaystyle \Rightarrow b=-6 , 2$$
Thus, the coordinates of the required points are $$(0, 2, 0)$$ and $$(0, -6,0)$$

By section formula,

$$\dfrac{2}{9} + \dfrac{2}{9} + \dfrac{2}{9} = \dfrac{6}{9}$$

Using Graph Paper

$$a(1,3,2) + b(1, - 5,6) + c(2,1, - 2) = (4,10, - 8)$$

The distance between the  points  $$(x_{1},y_{1},z_{1})$$ and  $$(x_{2},y_{2},z_{2})$$   is   $$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}$$

Given pints are $$(3,4)$$ and $$(6,1)$$

Given position of jewel box as original, $$O(0,0)$$ .

The distance between the  points  $$(x_{1},y_{1},z_{1})$$ and  $$(x_{2},y_{2},z_{2})$$   is   $$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}$$

Consider the problem 

Let us assume that point B divides the line AC in the ratio k:1

$$A=(2,-3,1),B(-6,5,3),C(8,7,-7)$$

According to question,

Consider the given point, $$R\left( 0,-3 \right),S\left( 0,\dfrac{5}{2} \right)$$

We know that distance between two points,

  $$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$$

  $$=\sqrt{{{\left( 0+3 \right)}^{2}}+{{\left( \dfrac{5}{2}-0 \right)}^{2}}}$$

$$=\sqrt{{{\left( 9\right)}}+{{\left( \dfrac{25}{4} \right)}}} =\dfrac{\sqrt61}{2}$$

$$A(x,\cfrac { 1 }{ x } ),B(\cfrac { 1 }{ x } ,x)$$

line AB is divided into three equal parts i.e, AC,CD,DB

Consider 

$$A(0,7,-7),\, B(1,4,-5), \, C(-1,10,-9)$$

$$A(7,6),B(3,4),O(\cfrac { m }{ 2 } ,0)$$

$$A=-3\hat { i } +3\hat { j } +5\hat { k } ,B=\hat { i } +2\hat { j } +3\hat { k } ,C=7\hat { i } -\hat { j } \\ \overrightarrow { AB } =4\hat { i } -\hat { j } -2\hat { k } ,\overrightarrow { AC } =10\hat { i } -4\hat { j } -5\hat { k }$$

If two points on $$x- axis$$ them $$,$$  

(i) The coordinates of point R that divides the line segment joining points $$P(x_1, y_1, z_1)$$ and $$Q(x_2, y_2, z_2)$$ internally in the ratio $$m:n$$ are
$$\left ( \dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}, \dfrac{mz_2 + nz_1}{m + n} \right )$$
Let $$R(x, y, z)$$ be the points that divides the line segment joining points $$(-2, 3, 5)$$ and $$(1, -4, 6)$$ internally in the ratio $$2 : 3$$
$$x = \dfrac{2(1) + 3(-2)}{2 + 3}, y = \dfrac{2(-4) + 3(3)}{2 + 3}$$ and $$z = \dfrac{2(6) + 3(5)}{2 + 3}$$
i.e., $$x = \dfrac{-4}{5}, y = \dfrac{1}{5},$$ and $$z = \dfrac{27}{5}$$
Thus, the coordinates of the required point are $$\left ( \dfrac{-4}{5}, \dfrac{1}{5}, \dfrac{27}{5} \right )$$
(ii) The coordinates of point R that divides the line segment joining points $$P(x_1, y_1, z_1)$$ and $$Q(x_2, y_2, z_2)$$ externally in the ratio $$m:n$$ are
$$\left ( \dfrac{mx_2 + nx_1}{m - n}, \dfrac{my_2 + ny_1}{m - n}, \dfrac{mz_2 + nz_1}{m - n} \right )$$
Let $$R(x, y, z)$$ be the point that divides the line segment joining points $$(-2, 3, 5)$$ and $$(1, -4, 6)$$ externally in the ratio $$2 : 3$$
$$x = \dfrac{2(1) - 3(2)}{2 - 3}, y = \dfrac{2(-4) - 3(3)}{2 - 3},$$ and $$z = \dfrac{2(6) - 3(5)}{2 - 3}$$
i.e., $$x = -8, y = 17,$$ and $$z = 3$$
Thus, the coordinates of the required point are $$(-8, 17, 3).$$

Let the required point on y-axis $$=P(0,y)$$
$$PA=\sqrt { { \left( 0-6 \right)  }^{ 2 }+{ \left( y-7 \right)  }^{ 2 } } =\sqrt { 36+{ y }^{ 2 }+49-14y } =\sqrt { { y }^{ 2 }-1y+85 }$$
$$PB=\sqrt { { \left( 0-4 \right)  }^{ 2 }+{ \left( y+3 \right)  }^{ 2 } } =\sqrt { 16+{ y }^{ 2 }+9+6y } =\sqrt { { y }^{ 2 }+6y+25 }$$
we have
$$\cfrac { PA }{ PB } =\cfrac { 1 }{ 2 } \Rightarrow \cfrac { { PA }^{ 2 } }{ { PA }^{ 2 } } =\cfrac { 1 }{ 4 }$$
$$\cfrac { { y }^{ 2 }-14y+85 }{ { y }^{ 2 }+6y+25 } =\cfrac { 1 }{ 4 } \Rightarrow 4{ y }^{ 2 }-56y+340={ y }^{ 2 }+6y+25\Rightarrow 3{ y }^{ 2 }-62y+315=0$$
$$\Rightarrow y=\cfrac { 62\pm \sqrt { 3844-3780 }  }{ 6 } =\cfrac { 62\pm 8 }{ 6 }$$
$$y=9,\cfrac { 35 }{ 3 }$$
So, the points on y-axis are $$(0,9)$$ $$(0,\cfrac{35}{3})$$

Given points $$A(2,1,5),B(3,4,3)$$

Given the point is $$(p,q,r)$$ .

The given point is (4,-3,0).

$$PQ=\sqrt { { \left( 5-2 \right)  }^{ 2 }+{ \left( 4-2 \right)  }^{ 2 } } =\sqrt { 9+4 } =\sqrt { 13 } =3.6055=3.61units\quad$$

It is known that the coordinates of the centroid of the triangle, whose vertices are
$$(x_1, y_1, z_1), (x_2, y_2, z_2)$$ and $$(x_3, y_3, z_3),$$ are
$$\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}, \dfrac{z_1 + z_2 + z_3}{3}$$
Therefore, coordinates of the centroid of
$$\Delta PQR = \left ( \dfrac{2a - 4 + 8}{3}, \dfrac{2 + 3b + 14}{3}, \dfrac{6 - 10 + 2c}{3} \right ) = \left ( \dfrac{2a + 4}{3}, \dfrac{3b + 16}{3}, \dfrac{2c - 4}{3} \right )$$
It is given that origin is the centroid of $$\Delta PQR$$
$$\Rightarrow \dfrac{2a + 3}{3} = 0, \dfrac{3b + 16}{3}$$ and $$\dfrac{2c - 4}{3} = 0$$
$$\Rightarrow a = -2b, b = \dfrac{16}{3}$$ and $$c = 2$$
Thus, the respective values of a, b and c are $$-2, -\dfrac{16}{3}$$ and $$2$$

Let A and B be the points that trisect the line segment joining points $$P(4, 2, -6)$$ and $$Q(10, -16, 6)$$
Point A divides PQ in the ratio $$1 :2$$ . Therefore, by section formula, the coordinates of point A are given by
$$\left ( \dfrac{1(10) + 2(4)}{1 + 2}, \dfrac{1(-16) + 2(2)}{1 + 2}, \dfrac{1(6) + 2(-4)}{1 + 2} \right ) = (6, -4, -2)$$ Point B divides PQ in the ratio $$2 : 1.$$ Therefore, by section formula, the coordinates of point B are given by
$$\left ( \dfrac{2(10) + 1(4)}{2 + 1}, \dfrac{2(-16) + 1(2)}{2 + 1}, \dfrac{2(6) + 1(-4)}{2 + 1} \right ) = (8, -10, 2)$$
Thus, $$(6, -4, -2)$$ and $$(8, -10, 2)$$ are the points that trisect the line segment joining points $$P(4, 2, -6)$$ and $$Q(10, -16, 6)$$

Let the YZ plane divide the line segment joining points $$(-2, 4, 7)$$ and $$(3, -5, 8)$$ in the ratio $$k :1.$$
Hence, by section formula, the coordinates of point of intersection are given by 
$$\left ( \dfrac{k(3) - 2}{k + 1}, \dfrac{k (-5) + 4}{k + 1}, \dfrac{k (8) + 7}{k + 1} \right )$$
On the YZ plane, the x-coordinate of any point is zero.
$$\Rightarrow \dfrac{3k - 2}{k + 1} = 0$$
$$\Rightarrow 3k - 2 = 0$$
$$\Rightarrow k = \dfrac{2}{3}$$
Thus, the YZ plane divides the line segment formed by joining the given points in the ratio $$2 : 3.$$

Let point $$Q(5, 4, -6)$$ divide the line segment joining points $$(3, 2, -4)$$ and $$R(9, 8, -10)$$ in the ratio $$k : 1.$$
Therefore, by section formula,
$$5, 4, -6) = \left ( \dfrac{k(9) + 3}{k + 1}, \dfrac{k(8) + 2}{k + 1}, \dfrac{k (-10) - 4}{k + 1} \right )$$
$$\Rightarrow \dfrac{9k + 3}{k + 1} = 5$$
$$\Rightarrow 9k + 3 = 5k + 5$$
$$\Rightarrow 4k = 2$$
$$\Rightarrow k = \dfrac{2}{4} = \dfrac{1}{2}$$
Thus, point Q divides PR in the ratio $$1 : 2.$$

Let $$A=(-1,1,3)$$ and $$B=(0,5,6)$$
Then application of distance formula gives us
$$AB=\sqrt{1^{2}+4^{2}+3^{2}}=\sqrt{26}=\sqrt{k}$$
Hence $$k=26$$ .
Similarly
$$C=(2,3,-1)$$ and $$C=(2,6,2)$$ . Then
$$CD=\sqrt{0^{2}+3^{2}+3^{2}}=3\sqrt{2}=3\sqrt{m}$$
Hence $$m=2$$ .
Therefore $$\dfrac{k}{m}=\dfrac{26}{2}=13$$ .

The planes forming the parallelopiped are $$x=-1,x=1;y=2,y=-1$$ and $$z=5,z=-1$$

Distance Between two position vectors $$ai+bj+ck$$ and $$xi+yj+zk$$ is given by
$$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } }$$
Given $$\sqrt { 2 }i$$ , $$-\sqrt { 3 }k$$
Distance $$=\sqrt { \left(\sqrt { 2 }-0 \right) ^{ 2 }+\left( 0-0 \right) ^{ 2 }+\left( 0+-\sqrt { 3 } \right) ^{ 2 } }$$
                $$=\sqrt { 5 }$$
            $$m^{2}=5$$

Let the line segment has a point $$R$$ intersecting on the surface of the sphere
$$\displaystyle \Rightarrow R=\left( \frac { 3\lambda  }{ 1+\lambda  } +\frac { 4\lambda +1 }{ 1+\lambda  } ,\frac { 5\lambda +2 }{ 1+\lambda  }  \right)$$  
So $${ 9\lambda  }^{ 2 }+{ 16\lambda  }^{ 2 }+1+8\lambda +{ 25\lambda  }^{ 2 }+4+20\lambda -25\left\{ { \lambda  }^{ 2 }+1+2\lambda  \right\} =0$$
$$\Rightarrow { 25\lambda  }^{ 2 }-22\lambda -20=0$$
$$\displaystyle \Rightarrow \lambda =\frac { 22\pm \sqrt { 484+2000 }  }{ 50 } =\frac { 22\pm \sqrt { 2484 }  }{ 50 } =\frac { 11\pm \sqrt { 621 }  }{ 25 }$$

(i) Let point $$(0, 7, -10), (1, 6, -6)$$ and $$(4, 9, -6)$$ be denoted by $$A, B$$ and $$C$$ respectively
$$\displaystyle AB=\sqrt{\left ( 1-0 \right )^{2}+\left ( 6-7 \right )^{2}+\left ( -6+10 \right )^{2}}$$
$$=\displaystyle \sqrt{\left ( 1 \right )^{2}+\left ( -1 \right )^{2}+\left ( 4 \right )^{2}}$$
$$=\displaystyle \sqrt{1+1+16}$$
$$=\displaystyle \sqrt{18}$$
$$\Rightarrow AB=\displaystyle 3\sqrt{2}$$
$$\displaystyle BC=\sqrt{\left ( 4-1 \right )^{2}+\left ( 9-6 \right )^{2}+\left ( -6+6 \right )^{2}}$$
= $$\displaystyle \sqrt{\left ( 3 \right )^{2}+\left ( 3 \right )^{2}}$$
= $$\displaystyle \sqrt{9+9}=\sqrt{18}$$
$$\Rightarrow BC=3\sqrt{2}$$
$$\displaystyle CA=\sqrt{\left ( 0-4 \right )^{2}+\left ( 7-9 \right )^{2}+\left ( -10+6 \right )^{2}}$$
= $$\displaystyle \sqrt{\left ( -4 \right )^{2}+\left ( -2 \right )^{2}+\left ( -4 \right )^{2}}$$
= $$\displaystyle \sqrt{16+4+16}=\sqrt{36}=6$$
Here $$AB = BC$$ $$\displaystyle \neq$$ $$CA$$
Thus the given points are the vertices of an isosceles triangle

(ii) Let $$(0, 7, 10), (-1, 6, 6)$$ and $$(-4, 9, 6)$$ be denoted by $$A, B$$ and $$C$$ respectively
$$\displaystyle AB =  \sqrt{\left ( -1-0 \right )^{2}+\left ( 6-7 \right )^{2}+\left ( 6-10 \right )^{2}}$$
$$\displaystyle =\sqrt{\left ( -1 \right )^{2}+\left ( -1 \right )^{2}+\left ( -4 \right )^{2}}$$
$$\displaystyle =\sqrt{1+1+16}=\sqrt{18}$$
$$\displaystyle =3\sqrt{2}$$    
$$\displaystyle BC=\sqrt{\left ( -4+1 \right )^{2}+\left ( 9-6 \right )^{2}+\left ( 6-6 \right )^{2}}$$
$$\displaystyle =\sqrt{\left ( -3 \right )^{2}+\left ( 3 \right )^{2}+\left ( 0 \right )^{2}}$$
$$\displaystyle =\sqrt{9+9}=\sqrt{18}$$
$$\displaystyle =3\sqrt{2}$$
$$\displaystyle CA=\sqrt{\left ( 0+4 \right )^{2}+\left ( 7-9 \right )^{2}+\left ( 10-6 \right )^{2}}$$
$$\displaystyle =\sqrt{\left ( 4 \right )^{2}+\left ( -2 \right )^{2}+\left ( 4 \right )^{2}}$$
$$\displaystyle =\sqrt{16+4+16}$$
$$\displaystyle =\sqrt{36}$$
$$=6$$
Now  $$\displaystyle AB^{2}+BC^{2}=\left ( 3\sqrt{2} \right )^{2}+\left ( 3\sqrt{2} \right )^{2}=18+18=36=AC^{2}$$
Therefore by pythagoras theorem $$ABC$$ is a right triangle
Hence the given points are the vertices of a right-angled triangle

(iii) Let $$(-1, 2, 1), (1, -2, 5), (4, -7, 8)$$ and $$(2, -3, 4)$$ be denoted by $$A, B, C$$ and $$D$$ respectively
$$\displaystyle AB=\sqrt{\left ( 1+1 \right )^{2}+\left ( -2-2 \right )^{2}+\left ( 5-1 \right )^{2}}$$
$$\displaystyle =\sqrt{4+16+16}$$
$$\displaystyle AB= \sqrt{36}$$
$$AB=6$$
$$BC=\sqrt { \left( 4-1 \right) ^{ 2 }+\left( -7+2 \right) ^{ 2 }+\left( 8-5 \right) ^{ 2 } }$$
$$\displaystyle =\sqrt{9+25+9}=\sqrt{43}$$
$$\displaystyle CD= \sqrt{\left ( 2-4 \right )^{2}+\left ( -3+7 \right )^{2}+\left ( 4-8 \right )^{2}}$$
$$\displaystyle =\sqrt{4+16+16}$$
$$\displaystyle =\sqrt{36}$$
$$CD=6$$
$$DA = \displaystyle \sqrt{\left ( -1-2 \right )^{2}+\left ( 2+3 \right )^{2}+\left ( 1-4 \right )^{2}}$$
$$\displaystyle DA=\sqrt{9+25+9}=\sqrt{43}$$
Here $$AB = CD = 6$$ , $$BC = AD = \displaystyle \sqrt{43}$$
Hence the opposite sides of quadrilateral $$ABCD$$ whose vertices are taken in order are equal
Therefore $$ABCD$$ is a parallelogram
Hence the given points are the vertices of a parallelogram  

Solution: Distance between two points $$(x_1, y_1)$$ ,
$$(x_2, y_2)$$ is  $$\sqrt{(x_2\, -\, x_1)^2\, +\, (y_2\, -\, y_1)^2}$$ .
Given, PQ = 5,
$$\Rightarrow\, \sqrt{(a\, -\, 11)^2\, +\, (1\, -\, (-2))^2}\, =\, 5$$
Taking square root on both sides, we get
$$(a\, -\, 11)^2\, =\, 25\, -\, 9\, =\,  16\, \Rightarrow\, a\, -\, 11$$
$$=\, \pm\, \sqrt{16}\, =\, a\, -\, 11\, =\, \pm\, 4$$
$$\Rightarrow\, a\, =\, 15\, or\, 7$$

Given points $$P(3, 2, -4), Q(5, 4, -6)$$ and $$R(9, 8, -10)$$  
Let $$Q$$ divides $$PR$$ in the ratio $$k:1$$ .
So, by section formula, coordinates of $$Q$$ are $$(\displaystyle \frac{k(9)+1(3)}{k+1}, \frac{k(8)+1(2)}{k+1}, \frac{k(-10)+1(-4)}{k+1}$$

$$=\displaystyle (\frac{9k+3}{k+1},\frac{8k+2}{k+1}, \frac{-10k+4}{k+1} )$$
But the given coordinates of $$Q$$ are $$(5,4,-6)$$ .
On comparing  $$\displaystyle \frac{9k+3}{k+1}=5$$
$$\Rightarrow 9k+3=5k+5$$
$$\Rightarrow \displaystyle  k=\frac{1}{2}$$
So, $$Q$$ divides $$PQ$$ in the ratio $$1:2$$ .

(i) The coordinates of point R that divides the line segment joining points $$P\displaystyle \left ( x_{1},y_{1},z_{1} \right )$$ and $$Q\displaystyle \left ( x_{2},y_{2},z_{2} \right )$$ internally in the ratio $$m:n$$ are

$$\displaystyle \left ( \frac{mx_{2}+nx_{1}}{m+n},\frac{my_{2}+ny_{1}}{m+n},\frac{mz_{2}+nz_{1}}{m+n} \right )$$

Let $$R (x, y, z)$$ be the point that divides the line segment joining points $$(-2, 3,5)$$ and $$(1, -4, 6)$$ internally in the ratio $$2:3$$
$$\displaystyle x=\frac{2\left ( 1 \right )+3\left ( -2 \right )}{2+3},y=\frac{2\left ( -4 \right )+3\left ( 3 \right )}{2+3}$$ and $$z=\displaystyle \frac{2\left ( 6 \right )+3\left ( 5 \right )}{2+3}$$

i.e.,  $$\displaystyle x=\frac{-4}{5},y=\frac{1}{5}$$ and $$z=\displaystyle \frac{27}{5}$$
Thus the coordinates of the required point are  $$\displaystyle \left ( -\frac{4}{5},\frac{1}{5},\frac{27}{5} \right )$$

(ii) The coordinates of point $$R$$ that divides the line segment joining points $$P \displaystyle \left ( x_{1},y_{1},z_{1} \right )$$ and $$Q \displaystyle \left ( x_{2},y_{2},z_{2} \right )$$ externally in the ratio $$m:n$$ are

$$\displaystyle \left ( \frac{mx_{2}-nx_{1}}{m-n},\frac{my_{2}-ny_{1}}{m-n}\frac{mz_{2}-nz_{1}}{m-n} \right )$$

Let $$R (x, y, z)$$ be the point that divides the line segment joining points $$(-2, 3, 5)$$ and $$(1, -4, 6)$$ externally in the ratio $$2:3$$
$$\displaystyle x=\frac{2\left ( 1 \right )-3\left ( -2 \right )}{2-3},y=\frac{2\left ( -4 \right )-3\left ( 3 \right )}{2-3}$$ and $$z=\displaystyle \frac{2\left ( 6 \right )-3\left ( 5 \right )}{2-3}$$
i.e., $$x=-8, y=17$$ and $$z=3$$
Thus the coordinates of the required point are $$(-8, 17, 3)$$

Let the coordinates of $$P$$ be $$(x, y, z)$$ .

Let $$P (x, y, z)$$ be the point that is equidistant from points $$A(1, 2, 3)$$ and $$B(3, 2, -1)$$ .

The equation of line $$AB$$ is given by $$\vec{r} = (3\hat{i} - 4\hat{j} - 5\hat{k}) + \lambda(\hat{i} - \hat{j} - 6\hat{k})$$

The given points are $$A (2, -3, 4), B (-1, 2, 1)$$ and  $$\displaystyle C\left ( 0,\frac{1}{3},2 \right )$$
Let $$P$$ be a point that divides $$AB$$ in the ratio $$k:1$$ .
Using section formula, the coordinates of $$P$$ are given by,
$$\displaystyle \left ( \frac{k\left ( -1 \right )+2}{k+1},\frac{k\left ( 2 \right )-3}{k+1} ,\frac{k\left ( 1 \right )+4}{k+1} \right )$$
Now, we will find the value of $$k$$ at which point $$P$$ coincides with point $$C$$ .
$$\Rightarrow \displaystyle \frac{-k+2}{k+1}=0$$ , we get  $$k=2$$
For $$k=2$$ , the coordinates of point $$P$$ are  $$\displaystyle \left ( 0,\frac{1}{3},2 \right )$$ ,
i.e.,  $$\displaystyle C\left ( 0,\frac{1}{3},2 \right )$$ is a point that divides $$AB$$ internally in the ratio $$2:1$$ and is the same as point $$P$$
Hence, points $$A, B$$ and $$C$$ are collinear.

Let the $$YZ$$ plane divide the line segment joining points $$(-2, 4, 7)$$ and $$(3, -5, 8)$$ in the ratio $$k:1$$ .

Let $$AD, BE$$ and $$CF$$ be the medians of the given $$\triangle ABC$$ .
Since $$AD$$ is the median, $$D$$ is the mid-point of $$BC$$ .

The coordinates of points $$P$$ and $$Q$$ are given as $$P (2, -3, 4)$$ and $$(8, 0, 10)$$

Given coordinates of points $$A$$ and $$B$$ are   $$(3, 4, 5)$$ and $$(-1, 3, -7)$$ respectively.

The line joining the points A and B is parallel to the vector pointing from A to B, hence its equation is:

$$\vec {PQ} = \vec Q - \vec P = -4\hat j+3\hat k$$

$$A = (1,1,1), B = (1,2,3), C = (2,3,3)$$

Let $$\vec{a},\vec{b},\vec{c}$$ be the position vectors of $$\Delta ABC$$ whose vertices are $$A(2,p,-3),B(q,-2,5),R(-5,1,r)$$

Given conic is  $$S =Ax^{2}+2Hxy+By^{2}-1=0$$

Let $$(x,y,z)$$ be a point with the sum of whose distance from $$(4,0,0)$$ and $$(-4,0,0)$$ is equal to 10. By using the distance formula between two points we have

see the above diagram

consider the equation of the line 

$$A = (1,2,3)$$
$$B = (-1,3,5)$$
$$A = i+2j+3k$$
$$B = -i+3j+5k$$
$$AB = B-A$$
$$AB = (-i+3j+5k) - (i+2j+3k)$$
$$AB = -2i+j+2k$$
$$AB = \sqrt{(-2)^2+(1)^2+(2)^2}=\sqrt{9}=3$$

$$\displaystyle DO=\sqrt { { \left( \frac { a }{ 2 } -0 \right)  }^{ 2 }+{ \left( \frac { b }{ 2 } -0 \right)  }^{ 2 } } =\frac { 1 }{ 2 } \sqrt { { a }^{ 2 }+{ b }^{ 2 } } ,$$
$$\displaystyle DA=\sqrt { { \left( \frac { a }{ 2 } -a \right)  }^{ 2 }+{ \left( \frac { b }{ 2 } -0 \right)  }^{ 2 } } =\frac { 1 }{ 2 } \sqrt { { a }^{ 2 }+{ b }^{ 2 } }$$

$$Let\>the\>point\>on\>X-Y\>plane\>be\>(a,b,0)\>and\>ratio\>of\>division\>be\>k:1\\then\>0\>=\>(\frac{-4k+5}{k+1})\\or\>k\>=\>(\frac{5}{4})=(\frac{m}{n})\\\therefore\>m+n\>=\>5+4\>=\>9$$

Plane: $$2x+y-z=3$$

Let three points are $$A,B,C$$

If the given points are vertices of a parallelogram then coordinates of mid point of $$AC$$ and $$BD$$ should be equal.

$$\begin{array}{l} \overrightarrow { AB } =\frac { 1 }{ 2 } \overrightarrow { CD } \, \, \, \, \, \, \, or\, \, \, \, \overrightarrow { AB } =-\frac { 1 }{ 2 } \overrightarrow { CD }  \\ \overrightarrow { AB } =-4\widehat { j } +\widehat { k }  \\ 2\left( { -4\widehat { j } +\widehat { k }  } \right) =\overrightarrow { D } -\overrightarrow { C } \, \, \, \, \, \, \, \, \, \, 2\left( { -4\widehat { j } +\widehat { k }  } \right) =\overrightarrow { C } -\overrightarrow { D }  \\ -8\widehat { j } +2\widehat { k } -\widehat { i } +2\widehat { j } +7\widehat { k } =\widehat { 0 } \, \, \, \, \, \, \, \, \, -8\widehat { j } +2\widehat { k } +\widehat { i } -2\widehat { j } -7\widehat { k } =\widehat { 0 }  \\ \overrightarrow { D } =-\widehat { i } -6\widehat { j } +9\widehat { k } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, -\widehat { i } -10\widehat { j } -5\widehat { k } =-\widehat { 0 } \,  \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, or\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \overrightarrow { D } =-\widehat { i } +10\widehat { j } +5\widehat { k } \,  \end{array}$$

Let the required point be $$P(x_1,y_1,z_1)$$

$$A(1,2,3),B(3,0,1)$$ and $$C(-1,1,-4)$$

Let the required point be $$P(x,y,z)$$

$$A(-5,4),B(-1,-2),C(5,2)\\ \overrightarrow { AB } =4\hat { i } -6\hat { j } ,\quad \overrightarrow { BC } =6\hat { i } +4\hat { j } ,\quad \overrightarrow { CA } =10\hat { i } -2\hat { j } \\ \left| \overrightarrow { AB }  \right| =\sqrt { { 4 }^{ 2 }+{ 6 }^{ 2 } } =2\sqrt { 13 } \\ \left| \overrightarrow { BC }  \right| =\sqrt { { 6 }^{ 2 }+{ 4 }^{ 5 } } =2\sqrt { 13 } \\ \left| \overrightarrow { CA }  \right| =\sqrt { { 10 }^{ 2 }+{ 2 }^{ 2 } } =2\sqrt { 26 }$$

(i) The coordinates of the point which divides the line joining the point (2,-4,3) and (-4,5,-6) in the ratio 1:-4 is given by

$${ { A } }\left( { { { 1,1,2 } } } \right) { { ,B } }\left( { { { 2,1,3 } } } \right) \, { { and } }\, { { C } }\left( { { { 1,3,5 } } } \right)$$

Given that, points $$P(3,2,-4),\,Q(5,4,-6)$$ and $$R(9,8,-10)$$ are collinear.

$${X_1},{Y_1},{Z_1}$$ = $$2,-1,3$$

we have,

$$P(x,y,z)$$

$$\dfrac{x+1}{2}=\dfrac{y+2}{3}=\dfrac{z-3}{6}$$

$$\begin{array}{l} In\, \, the\, \, \Delta ABC \\ B{ C^{ 2 } }=A{ C^{ 2 } }+A{ B^{ 2 } } \\ { \left( { 3-x } \right) ^{ 2 } }+4+36=1+1+25+{ \left( { 4-x } \right) ^{ 2 } }+9+1 \\ 9+6x+{ x^{ 2 } }+40=37+16-8x+{ x^{ 2 } } \\ 2x=-49+37+16 \\ =-49+53 \\ =4 \\ x=2 \end{array}$$

We have,

Let $$AB$$ be the line segment joining the points

$$A\left( 1,2,3 \right)$$ and $$B\left( -3,4,-5 \right)$$ .

Let $$XY-plane$$ divide line $$AB$$ .at $$P\left( x,y,z \right)$$ in the ratio $$k:1$$ .

Co- ordinate of point $$P\left( x,y,z \right)$$ that divide line segment joining point $$A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$$ and $$B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$$ in the ratio $$m:n$$ is

$$=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\,\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\,\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)$$

Here,

$$m:n=k:1$$

$$A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 1,2,3 \right)$$

$$B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( -3,4,-5 \right)$$

Then the co-ordinate of

$$P\left( x,y,z \right)=\left( \dfrac{k\times \left( -3 \right)+1\left( 1 \right)}{k+1},\dfrac{k\times \left( 4 \right)+1\times \left( 2 \right)}{k+1},\dfrac{k\times \left( -5 \right)+1\times \left( 3 \right)}{k+1} \right)$$

$$=\left( \dfrac{-3k+1}{k+1},\dfrac{4k+2}{k+1},\dfrac{-5k+3}{k+1} \right)$$

Since, point $$P\left( x,y,z \right)$$ lie on the $$XY-plane$$

Then, Its Z-coordinate will be zero.

$$P\left( x,y,0 \right)=\left( \dfrac{-3k+1}{k+1},\dfrac{4k+2}{k+1},\dfrac{-5k+3}{k+1} \right)$$

Then, Comparing and we get,

$$\dfrac{-5k+3}{k+1}=0$$

$$-5k+3=0$$

$$-5k=-3$$

$$k=\dfrac{3}{5}$$

Then, $$k:1=3:5$$

We can substitute the value of k in (1), we get,

$$= (\dfrac {−0.8}{1.6}, \dfrac {4.4}{1.6},0)$$

$$= (-0.5, 2.75, 0)$$  

We get the positive vector of the point of division from (2). Thus, the vector is given by $$-0.5 \hat i + 2.75 \hat j.$$ Here, $$\hat i$$ denotes unit vector to X-axis and $$\hat j$$ denotes unit vector to Y-axis.

Hence, this is the answer.

$$\begin{array}{l} \frac { { x-8 } }{ 3 } =\frac { { y+9 } }{ { -16 } } =\frac { { z-10 } }{ 7 }  \\ \frac { { x-15 } }{ 3 } =\frac { { y-29 } }{ 8 } =\frac { { z-5 } }{ { -5 } }  \\ d=\left| { \frac { { \left\{ { \left( { 3\widehat { i } -16\widehat { j } +7\widehat { k }  } \right) \times \left( { 3\widehat { i } +8\widehat { j } -5\widehat { k }  } \right)  } \right\} .\left( { 7\widehat { i } +38\widehat { j } -5\widehat { k }  } \right)  } }{ { \left| { \left( { 3\widehat { i } -16\widehat { j } +7\widehat { k }  } \right) \times \left( { 3\widehat { i } +8\widehat { j } -5\widehat { k }  } \right)  } \right|  } }  } \right|  \\ =\left| { \frac { { \left( { 24\widehat { i } +36\widehat { j } +72\widehat { k }  } \right) \times \left( { 7\widehat { i } +38\widehat { j } -5\widehat { k }  } \right)  } }{ { \left| { \left( { 24\widehat { i } +36\widehat { j } +72\widehat { k }  } \right)  } \right|  } }  } \right|  \\ =\left| { \frac { { 168+1368-360 } }{ { \sqrt { 576+1296+5184 }  } }  } \right|  \\ =\frac { { 1176 } }{ { \sqrt { 7056 }  } }  \\ =\frac { { 1176 } }{ { 84 } }  \\ =14\, \, unit. \end{array}$$

Given : 

Let the ratio be $$m:n$$

Given that 

Distance of $$(p,q,r)$$ from $$x$$ -axis is $$=p$$ units.

The equation of plane= $$\dfrac{x}{1}+\dfrac{y}{2}+\dfrac{z}{3}=1=6x+3y+2z=6$$

Let $$P(x, y)$$ divide the line $$A(-2, 5)$$ and $$B(3, -5)$$ in the ratio $$2:3$$ .

$$A\left( { -5,7 } \right) \, \, \, B\left( { -1,3 } \right)  \\ AB=\sqrt { { { \left( { -5-\left( { -1 } \right)  } \right)  }^{ 2 } }+{ { \left( { 7-3 } \right)  }^{ 2 } } }  \\ \, \, \, \, \, \, \, \, \, =\sqrt { 16+16 }  \\ \, \, \, \, \, \, \, \, \, =4\sqrt { 2 } \, \, units$$

$$\begin{array}{l} A\left( { { x_{ 1 } },{ y_{ 1 } } } \right) =A\left( { 2,-3 } \right)  \\ B\left( { { x_{ 2 } },{ y_{ 2 } } } \right) =B\left( { 3,2 } \right)  \\ AB=\sqrt { { { \left( { { x_{ 1 } }-{ x_{ 2 } } } \right)  }^{ 2 } }+{ { \left( { { y_{ 1 } }-{ y_{ 2 } } } \right)  }^{ 2 } } }  \\ AB=\sqrt { { { \left( { -2-3 } \right)  }^{ 2 } }+{ { \left( { -3-2 } \right)  }^{ 2 } } }  \\ =5\sqrt { 2 } \, \, \, \, \, \, \, units. \end{array}$$

The points of Tetraheadron are  $$(3,4,5),(2,5,9),(5,2,8),(2,5,2)$$  

$$\textbf{Hint:  Suppose}$$   $${\rm{C}}$$   $$\textbf{divide the line}$$   $${\rm{AB}}$$   $$\textbf{in the ratio,}$$   $${\rm{k}}:1$$

Solution:

$$\textbf{Step 1: Use section formula}$$

Given points are   $${\rm{A}}\left( { - 1,2} \right),{\rm{C}}\left( {2,5} \right),{\rm{B}}\left( {5,8} \right)$$

 The coordinate of point  which divide the line segment joining the point $$\left(x_1,y_1\right)$$ and $$\left(x_2, y_2\right)$$ in the ratio   $$m:n$$ is given by $$\left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$$

$$\therefore \left( {\dfrac{{{\rm{m}}{{\rm{x}}_2} + {\rm{n}}{{\rm{\ x}}_1}}}{{{\rm{\;m}} + {\rm{n}}}},\dfrac{{{\rm{m}}{{\rm{y}}_2} + {\rm{n}}{{\rm{y}}_1}}}{{{\rm{\;m}} + {\rm{n}}}}} \right) = \left( {2,5} \right)$$

Let $${\rm{C}}$$ divide the line $${\rm{AB}}$$ in the ration, $${\rm{k}}:1$$

$$\therefore \left( {\dfrac{{{\rm{k}}\left( 5 \right) + 1\left( { - 1} \right)}}{{{\rm{k}} + 1}},\dfrac{{{\rm{k}}\left( 8 \right) + 1\left( 2 \right)}}{{{\rm{k}} + 1}}} \right) = \left( {2,5} \right)$$

$$\Rightarrow \dfrac{{5{\rm{k}} - 1}}{{{\rm{k}} + 1}} = 2 \ldots ..\left( 1 \right)$$

$$\Rightarrow \dfrac{{8{\rm{k}} + 2}}{{{\rm{k}} + 1}} = 5 \ldots  \ldots .\left( 2 \right)$$

$$\textbf{Step 2: Solve the above equation}$$

Solving equation (1), we get

$$5{\rm{k}} - 1 = 2{\rm{k}} + 2$$

$$\begin{array}{l} \Rightarrow 3k = 3\\ \Rightarrow k = 1\end{array}$$

Solving equation (2), we get

$$\begin{array}{l}8k + 2 = 5k + 5\\\; \Rightarrow 3k = 3\\ \Rightarrow k = 1\end{array}$$

∴ C divides the line segment joining A and B in the ratio $$1:1$$

$$\Rightarrow {\rm{C}}$$ is the mid-point of line $${\rm{AB}}$$

Therefore, $${\rm{A}}\left( { - 1,2} \right),{\rm{C}}\left( {2,5} \right),{\rm{B}}\left( {5,8} \right)$$

$$\textbf{Hence, Proved}$$

The distance between the points $$(-1,5,-5)$$ and $$(0,0,0)$$ is $$\sqrt{(-1-0)^2+(5-0)^2+(-5-0)^2}$$

$$(h-1)^2+(k+2)^2+(l-3)^2\\=(h+3)^2+(k-4)^2+(l-2)^2\\ \Rightarrow -2h+4k-6l+14\\ \Rightarrow 6h-8k-4l+29\\ \Rightarrow 8x-12y+2z+15=0$$

$${\textbf{Step -1: Define distance formula}}$$

                  $${\text{Origin is (0,0) and given point is ( 3,4)}}$$

                  $${\text{Distance of point ( 3,4) from the origin = Distance of point ( 3,4) from (0,0)}}$$

                  $${\text{Distance formula}} = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$$

$${\textbf{Step -2: Find distance}}$$

                  $$({x_1},{y_1}) = ( 3,4) {\text{ and (}}{x_2},{y_2}) = (0,0)$$

                  $$= \sqrt {{{(0 - 3)}^2} + {{(0 - 4)}^2}}$$

                  $$= \sqrt {9 + 16}$$

                  $$= \sqrt {25}$$

                  $$= 5\;{\text{units}}$$

$${\textbf{ Hence, the distance is 5 units}}{\text{.}}$$  

$$(a)$$

All three are positive, hence (5,2,3) lie in 1st octant

At y axis,

$$(h,k,l)\\2\sqrt{(h+2)^2+(k-3)^2+(l-4)^2}\\=\sqrt{(h-3)^2+(k+1)^2+(l+2)^2}\\ \Rightarrow 4(h^2+4+4h+k^2+9-6k+l^2+16-8l)=(h^2+k^2+l^2+9+1+4-6h+2k+4l)\\ \Rightarrow 3h^2+3k^2+3l^2+22h-26k-36l+102=0\\ \Rightarrow 3x^2+3y^2+3z^2+22x-26y-36z+102=0$$

So correct answer 1.

Let the vertices of parallelogram be,

Let $$O (\alpha, \beta , \gamma)$$ be the image of the point $$P(3, 2, 1)$$ in the plane $$2x - y  + z + 1 =0$$

Any point in the $$yz -$$ plane is of the form $$P(0, y, z)$$ .

Given : Point $$A (2, 1, -3)$$ and $$B (5, -8, 3)$$

Given: Three vertices $$A (3, 4, -1), B(7, 10, -3)$$ and $$C(5, -2, 7)$$ of a parallelogram ABCD.

The equation of line passing through the points $$(3, -4, -5)$$ and $$(2, -3, 1)$$ is
Let the line (i) crosses the plane $$2x + y + z = 7$$ at point $$P (\alpha, \beta, \gamma)$$ ...(ii)
Also $$P(\alpha, \beta, \gamma)$$ lie on plane (ii)
$$\therefore 2\alpha + \beta + \gamma = 7$$
$$\Rightarrow 2(-\lambda + 3) + (\lambda - 4) + (6\lambda - 5) = 7$$
$$\Rightarrow -2\lambda + 6 + \lambda - 4 + 6\lambda - 5 = 7$$
Hence, the coordinate of required point $$P$$ is $$(-2 + 3, 2 - 4, 6 \times 2 - 5) \,i.e., (1, -2, 7)$$

Let coordinates of D be (x, y, z)
Direction vector along AB is
$$\therefore$$ Equation of line AB, is given by
Direction vector along BC is
$$\therefore$$ Equation of a line BC, is given by
Since ABCD is a parallelogram AC and BD bisect each other
$$\therefore \left [ \dfrac{4 + 1}{2}, \dfrac{5 + 2}{2}, \dfrac{10 - 1}{2} \right ] = \left [ \dfrac{2 + x}{2}, \dfrac{3 + y}{2}, \dfrac{4 + z}{2} \right ]$$
$$\Rightarrow 2 + x = 5, 3 + y = 7, 4 + z = 9$$
Coordinates of D are $$(3, 4, 5).$$

Vertices of $$\delta ABC$$ are $$A(-8 , 4) , B(-6 , 6)$$ and $$C(-3 , 9)$$ . 
$$\therefore \,$$ Area of $$\Delta ABC = \dfrac{1}{2} [x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3(y_1 - y_2)]$$  

$$\Rightarrow \,$$ Area of $$\Delta = \dfrac{1}{2} [-8 (6 - 9) - 6 (9 - 4) - 3(4 - 6)]$$  

$$= \dfrac{1}{2} [-8(-3) - 6(5) - 3(-2)]$$

$$= \dfrac{1}{2} [24 - 30 + 6] = 0$$  
Hence , the area of given triangle is zero. 

Let $$A=(1,1,0),B=(1,0,1),C=(0,1,1)$$

Let $$G=(1,2,-1)$$ be the centroid of the tetrahedron $$OABC$$ .

The position vectors $$\bar {a},\ \bar {b},\ \bar {c}$$ of the points $$L,\ M,\ N$$ are
$$\bar {a} = 3\hat {i}-2\hat {j}-3\hat {k},\ \bar {b} = 7\hat {i} + \hat {k},\ \bar {c} = \hat {i}+2\hat {j}+\hat {k}$$
$$\overline {LM} = \bar {b}-\bar {a}=(7\hat {i}+\hat {k}) - (3\hat {i} - 2\hat {j} -3\hat {k}) = 4\hat {i}+2\hat {j}+4\hat {k}$$
$$\overline {MN} = \bar {c}-\bar {b}=(\hat {i}+2\hat {j}+\hat {k}) - (7\hat {i} + \hat {k}) = -6\hat {i}+2\hat {k}$$
$$\overline {NL} = \bar {a}-\bar {c}=(3\hat {i}-2\hat {j}-3\hat {k}) - (\hat {i} + 2\hat {j}+\hat {k}) = 2\hat {i}-4\hat {j}-4\hat {k}$$
$$\therefore l(LM) = \left| \overline { AB }  \right| = \sqrt {4^2+2^2+4^2} = \sqrt {16+4+16} = \sqrt {36} = 6$$
$$\therefore l(MN) = \left| \overline { MN }  \right| = \sqrt {(-6)^2+2^2} = \sqrt {36+4} = \sqrt {40} = \sqrt {10 \times 4} = 2\sqrt {10}$$
$$\therefore l(NL) = \left| \overline { NL }  \right| = \sqrt {(2)^2+(-4)^2+(-4)^2} = \sqrt {4+16+16} = \sqrt {36} = 6$$
$$\therefore l(LM) = 6,\ l(MN)=2\sqrt {10},\ l(NL)=6$$
$$\therefore \triangle LMN$$ is isosceles.

The position vectors $$\bar {a},\ \bar {b},\ \bar {c}$$ of the points $$A,\ B,\ C$$ are
$$\bar {a} = 2\hat {i}-\hat {j},\ \bar {b} = 4\hat {i} + \hat {j} + \hat {k},\ \bar {c} = 4\hat {i}-5\hat {j}+4\hat {k}$$
$$\overline {AB} = \bar {b}-\bar {a}=(4\hat {i}+\hat {j}+\hat {k}) - (2\hat {i} - \hat {j}) = 2\hat {i}+2\hat {j}+\hat {k}$$
$$\overline {BC} = \bar {c}-\bar {b}=(4\hat {i}-5\hat {j}+4\hat {k}) - (4\hat {i} + \hat {j} + \hat {k}) = -6\hat {j}+3\hat {k}$$
$$\overline {CA} = \bar {a}-\bar {c}=(2\hat {i}-\hat {j}) - (4\hat {i} - 5\hat {j}+4\hat {k}) = -2\hat {i}+4\hat {j}-4\hat {k}$$
$$\therefore l(AB) = \left| \overline { AB }  \right| = \sqrt {2^2+2^2+1^2} = \sqrt {4+4+1} = \sqrt {3} = 3$$
$$\therefore l(BC) = \left| \overline { BC }  \right| = \sqrt {(-6)^2+3^2} = \sqrt {36+9} = \sqrt {45} = 3\sqrt {5}$$
$$\therefore l(CA) = \left| \overline { CA }  \right| = \sqrt {(-2)^2+4^2+(-4)^2} = \sqrt {4+16+16} = \sqrt {36} = 6$$
$$\therefore { \left[ l\left( AB \right)  \right]  }^{ 2 }+{ \left[ l\left( CA \right)  \right]  }^{ 2 } = 3^2 + 6^2 = 9 + 36 = 45 = \left(3\sqrt {5}\right)^{2}$$
$$= \left[ l( BC )  \right]  ^{ 2 }$$
$$\therefore \triangle ABC$$ is right-angled at $$A$$ .

Let $$\bar{a}, \bar{b}, \bar{c}$$ be the position vectors of $$A, B$$ and $$C$$ respectively.

Let $$\bar{p}, \bar{l}, \bar{m}, \bar{n}$$ be the position vectors of the points $$K, L, M, N$$ respectively w.r.t. the origin $$O$$ .

Let $$\bar {a},\ \bar {b},\ \bar {c},\ \bar {d}$$ be the position vectors of $$A,\ B,\ C,\ D$$ respectively w.r.t the origin $$O.$$
Then $$\bar {a} = 3\hat {i}+2\hat {j}-\hat {k},\ \bar {b} = -2\hat {i}+2\hat {j}-3\hat {k},\ \bar {c} = 3\hat {i}+5\hat {j}-2\hat {k},\ \bar {d} = -2\hat {i}+5\hat {j}-4\hat {k}$$
$$\therefore \overline {AB} = \bar {b} - \bar {a}$$
$$=(-2\hat {i}+5\hat {j}-4\hat {k})-(3\hat {i}+2\hat {j}-\hat {k})$$
$$-5\hat {i}-2\hat {k}$$
$$\therefore \overline {DC} = \bar {c} - \bar {d}$$
$$= (3\hat {i}+5\hat {j}-2\hat {k}) - (-2\hat {i}+5\hat {j}-4\hat {k})$$
$$-5\hat {i}+2\hat {k}$$
$$=-(-5\hat {i}-2\hat {k})$$
$$\therefore \overline {DC} = \overline {AB}$$
$$\therefore \overline {DC}$$ is scalar multiple of $$\overline {AB}$$
$$\therefore \overline {DC}$$ is parallel $$\overline {AB}$$
Also, $$\left|DC \right| = \sqrt {5^2+2^2} =\sqrt{25+4}=\sqrt {29}$$
and $$\left|AB \right| = \sqrt {(-5)^2+(-2)^2} =\sqrt{25+4}=\sqrt {29}$$
$$\therefore \left|DC = \right| \left|AB \right|$$
$$\therefore l(AB) = l(DC)$$
$$\therefore$$ opposite sides $$AB$$ and $$DC$$ of $$ABCD$$ are aprallel and equal.
$$\therefore ABCD$$ is a parallelogram.

Let $$\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}=\lambda$$

Let $$O$$ is origin.
$$\therefore$$ Position vector of points $$A, B$$ and $$C$$ are respectively.
$$\vec{OA}=\vec{a}=4\hat{i}+5\hat{j}+10\hat{k}$$
$$\vec{OB}=\vec{b}=2\hat{i}+3\hat{j}+4\hat{k}$$
$$\vec{OC}=\vec{c}=\hat{i}+2\hat{j}-\hat{k}$$
(i) $$P(x, y, z)$$ is any point of line $$AB$$ whose position vector is $$\vec{r}$$ , then vector equation of line.
$$\vec{r}=\vec{a}+\lambda (\vec{b}-\vec{a})$$         $$(\because \vec{r}=x\hat{i}+y\hat{j}+z\hat{k})$$
$$\therefore \vec{r}=4\hat{i}+5\hat{j}+10\hat{k}+\lambda \left\{2\hat{i}+3\hat{j}+4\hat{k}-(4\hat{i}+5\hat{j}+10\hat{k})\right\}$$
$$\Rightarrow \vec{r}=4\hat{i}+5\hat{j}+10\hat{k}+\lambda (-2\hat{i}-2\hat{j}-6\hat{k})$$       $$.....(1)$$  
For cartesian equation of line $$AB$$ .
Put $$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$ in equation $$(1)$$
$$x\hat{i}+y\hat{j}+z\hat{k}=4\hat{i}+5\hat{j}+10\hat{k}+\lambda (-2\hat{i}-2\hat{j}-6\hat{k})$$
$$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(4-2\lambda)\hat{i}+(5-2\lambda) \hat{j}+(10-6\lambda)\hat{k}$$
Comparing the coefficients of $$\hat{i}, \hat{j}$$ and $$\hat{k}$$ both sides
$$x=4-2\lambda, y=5-2\lambda, z=10-6\lambda$$
$$\Rightarrow \dfrac{x-4}{-2}=\lambda, \dfrac{y-5}{-2}=\lambda, \dfrac{z-10}{-6}=\lambda$$
$$\Rightarrow \dfrac{x-4}{-2}=\dfrac{y-5}{-2}=\dfrac{z-10}{-6}=\lambda$$
$$\Rightarrow \dfrac{x-4}{1}=\dfrac{y-5}{1}=\dfrac{z-10}{3}$$
Remember: Line $$AB$$ passes through $$A(4,5,10)$$ and $$B(2,3,4)$$ .
$$\therefore$$ Position vector of $$A$$
$$\vec{a}=4\hat{i}+5\hat{j}+10\hat{k}$$
and $$\vec{b}=2\hat{i}+3\hat{j}+4\hat{k}$$
Then $$\vec{r}=\vec{a}+\lambda (\vec{b}-\vec{a})$$
(ii) For line $$BC$$
$$BC$$ passes through points $$B(2,3,4)$$ and $$C(1,2,-1)$$ .
$$=\vec{b}=2\hat{i}+3\hat{j}+4\hat{k}$$
Position vector of $$C=\vec{c}=\hat{i}+2\hat{j}-\hat{k}$$
$$\therefore$$ Vector equation of line $$BC$$
$$\vec{r}=\vec{b}+\mu (\vec{c}-\vec{b})$$
$$\Rightarrow \vec{r}=2\hat{i}+3\hat{j}+4\hat{k}+\mu \left\{\hat{i}+2\hat{j}-\hat{k}-(2\hat{i}+3\hat{j}+4\hat{k})\right\}$$
$$\Rightarrow \vec{r}=2\hat{i}+3\hat{j}+4\hat{k}+\mu (-\hat{i}-\hat{j}-5\hat{k})$$     $$....(2)$$
For cartesian equation of line $$BC$$ .
Let $$Q(x,yz)$$ which position vector is
$$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$
On putting the value of $$\vec{r}$$ in eq. $$(2)$$
$$x\hat{i}+y\hat{j}+z\hat{k}=2\hat{i}+3\hat{j}+4\hat{k}+\mu (-\hat{i}-\hat{j}-5\hat{k})$$
$$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(2-\mu)\hat{i}+(3-\mu)\hat{j}+(4-5\mu)\hat{k}$$
On comparing the coefficients of $$\hat{i}, \hat{j}, \hat{k}$$ in both sides
$$x=2-\mu, y=3-\mu, z=4-5\mu$$
$$\Rightarrow \dfrac{x-2}{-1}=\mu, \dfrac{y-3}{-1}=\mu, \dfrac{z-4}{-5}=\mu$$
$$\Rightarrow \dfrac{x-2}{1}=\dfrac{y-3}{1}=\dfrac{z-4}{5}=\mu$$
$$\Rightarrow \dfrac{x-2}{1}=\dfrac{y-3}{1}=\dfrac{z-4}{5}$$
(iii) For coordinates of point $$D$$ .
Let $$(x_{1}, y_{1}, z_{1})$$ are the coordinates of point $$D$$ .
$$\because ABCD$$ is a parallelogram whose diagonals $$AC$$ and $$BD$$ bisect each other at point $$D$$ . Therefore mid points of $$AC$$ and $$BD$$ will coincide. So the coordinate of mid point of $$AC$$ .
$$=\dfrac{4+1}{2}, \dfrac{5+2}{2}, \dfrac{10-1}{2}$$
$$=\dfrac{5}{2}, \dfrac{7}{2}, \dfrac{9}{2}$$
$$\therefore$$ Mid point of $$AC$$ is $$P\left(\dfrac{5}{2}, \dfrac{7}{2}, \dfrac{9}{2}\right)$$
Again coordinate of mid point $$Q$$ .
$$=\dfrac{2+x_{1}}{2}, \dfrac{3+y_{1}}{2}, \dfrac{4+z_{1}}{2}$$
or mid point of $$BD$$ is
$$Q\left(\dfrac{2+x_{1}}{2}, \dfrac{3+y_{1}}{2}, \dfrac{4+z_{1}}{2}\right)$$
$$\because$$ Mid points of $$AC$$ and $$BD$$ coincide.
$$\therefore \dfrac{5}{2}=\dfrac{2+x_{1}}{2}, \dfrac{7}{2}=\dfrac{3+y_{1}}{2}, \dfrac{9}{2}=\dfrac{4+z_{1}}{2}$$
$$\Rightarrow x_{1}=5-2, y_{1}=7-3, z_{1}=9-4$$
$$\Rightarrow x_{1}=3, y_{1}=4, z_{1}=5$$
$$\therefore (x_1, y_1, z_1)=(3,4,5)$$
$$\therefore$$ Coordinate of point $$D=(3,4,5)$$ .

Let the plane $$ax+by+cz+d=0$$ divides the line joining $$\left( { x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 } \right)$$ and $$\left( { x }_{ 2 },{ y }_{ 2 },{ z }_{ 2 } \right)$$ in the ratio $$k:1$$
$$\therefore$$ coordinates of $$\displaystyle p\left( \frac { k{ x }_{ 2 }+{ x }_{ 1 } }{ k+1 } ,\frac { { ky }_{ 2 }+{ y }_{ 1 } }{ k+1 } ,\frac { { kz }_{ 2 }+{ z }_{ 1 } }{ k+1 }  \right)$$  must satisfy $$\displaystyle ax+by+cz+d=0$$
i.e., $$\displaystyle a\left( \frac { { kx }_{ 2 }+{ x }_{ 1 } }{ k+1 }  \right) +b\left( \frac { { ky }_{ 2 }+{ y }_{ 1 } }{ k+1 }  \right) +c\left( \frac { { kz }_{ 2 }+{ z }_{ 1 } }{ k+1 }  \right) +d=0$$
$$\Rightarrow a\left( { kx }_{ 2 }+{ x }_{ 1 } \right) +b\left( { ky }_{ 2 }+{ y }_{ 1 } \right) +c\left( { kz }_{ 2 }+{ z }_{ 1 } \right) +d\left( k+1 \right) =0$$
$$\displaystyle \Rightarrow k\left( { ax }_{ 2 }+{ by }_{ 2 }+{ cz }_{ 2 }+d \right) +\left( { ax }_{ 1 }+{ by }_{ 1 }+{ cz }_{ 1 }+d \right) =0$$

On $$y-z$$ plane, $$x=0$$