Explanation
$$A(a,b,3)\\B(2,0,1)\\C(1,-1,-3)\\if\>A,Band\>C\>are\>collinear\>,then\\\>\vec{AB}\>\>=\lambda\>\vec{BC}\\\vec{AB}\>(2-a)\hat{i}-b\hat{j}-4\hat{k}\\\vec{BC}=-\hat{i}-\hat{j}-2\hat{k}\\now\>2-a=\lambda\>(-1)\\or\>2-a=-\lambda\>\rightarrow\>(1.)\\\>and\>-b=\>\lambda\>(-1)\lambda\>\\or\>-b=-\lambda\>\rightarrow\>(2.)\\\therefore\>from\>eq^n\>(1.)and(2.)\\2-a=-b\\or\>a-b=\>2\\now\>-4=\lambda\>(-2)\\\therefore\>\lambda\>=2\\then\>b=\lambda\>=2\\and\>a=2+\lambda\>=2+2=4$$
As all the points lie on the same line this implies that area of the triangle formed by the points is zero.
So,using area formula in determinant form
$${ X }_{ 1 }∗({ Y }_{ 2 }−{ Y }_{ 3 })−{ X }_{ 2 }∗({ Y }_{ 1 }−{ Y }_{ 3 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$$
$${ X }_{ 1 }∗({ Y }_{ 2 }−{ Y }_{ 3 })+{ X }_{ 2 }∗({ Y }_{ 3 }−{ Y }_{ 1 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$$
Now,divide both sides by $$ { X }_{ 1 }∗{ X }_{ 2 }∗{ X }_{ 3 }$$
$$({ Y }_{ 2 }−{ Y }_{ 3 })+{ X }_{ 2 }∗({ Y }_{ 3 }−{ Y }_{ 1 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$$
$$\dfrac { { Y }_{ 2 }−{ Y }_{ 3 } }{ { X }_{ 2 }−{ X }_{ 3 } } +\dfrac { { Y }_{ 3 }−{ Y }_{ 1 } }{ { X }_{ 3 }−{ X }_{ 1 } } +\dfrac { { Y }_{ 1 }−{ Y }_{ 2 } }{ { X }_{ 1 }−{ X }_{ 2 } } =0$$
Its collinear.
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