Explanation
$$\left (\dfrac {3}{2}, \dfrac {-3}{2}, \dfrac {7}{4}\right )$$
$${\textbf{Step - 2: Taking y co-ordinate equals to zero}}$$
$${\text{Since, this points lies in XOZ plane then its y co-ordinate should be zero}}$$
$$ \Rightarrow$$ $$\dfrac{{3\lambda - 1}}{{\lambda + 1}} = 0$$
$$ \Rightarrow$$ $${{3\lambda - 1}}= 0$$
$$ \Rightarrow$$ $$3\lambda= 1$$
$$\Rightarrow$$ $$\lambda = \dfrac{1}{3}$$
Since $$\overrightarrow { OP } $$ has projection $$\displaystyle \dfrac { 13 }{ 5 } ,\dfrac { 19 }{ 5 } $$ and $$\displaystyle \dfrac { 26 }{ 5 } $$ on the coordinate axes, therefore $$\displaystyle \overrightarrow { OP } =\dfrac { 13 }{ 5 } i+\dfrac { 19 }{ 5 } j+\dfrac { 26 }{ 5 } k$$
Let $$P$$ divides the join of $$Q\left( 2,2,4 \right) $$ and $$R\left( 3,5,6 \right) $$ in the ratio $$\lambda :1$$.
The position vector of $$P$$ is $$\displaystyle \left( \dfrac { 3\lambda +2 }{ \lambda +1 } \right) i+\left( \dfrac { 5\lambda +2 }{ \lambda +1 } \right) j+\left( \dfrac { 6\lambda +4 }{ \lambda +1 } \right) k$$
$$\displaystyle \therefore \dfrac { 13 }{ 5 } i+\dfrac { 19 }{ 5 } j+\dfrac { 26 }{ 5 } k=\left( \dfrac { 3\lambda +2 }{ \lambda +1 } \right) i+\left( \dfrac { 5\lambda +2 }{ \lambda +1 } \right) j+\left( \dfrac { 6\lambda +4 }{ \lambda +1 } \right) k$$
$$\displaystyle \Rightarrow \dfrac { 3\lambda +2 }{ \lambda +1 } =\dfrac { 13 }{ 5 } ,\dfrac { 5\lambda +2 }{ \lambda +1 } =\dfrac { 19 }{ 5 } ,\dfrac { 6\lambda +4 }{ \lambda +1 } =\dfrac { 26 }{ 5 }$$
$$ \Rightarrow \lambda =\dfrac { 3 }{ 2 } $$
Equation of the line through $$\left( 1,-2,3 \right) $$ parallel to the line $$\displaystyle \dfrac { x }{ 2 } =\dfrac { y }{ 3 } =\dfrac { z-1 }{ -6 } $$ is
$$\displaystyle \dfrac { x-1 }{ 2 } =\dfrac { y+2 }{ 3 } =\dfrac { z-1 }{ -6 } =r$$ (say) ...$$(1)$$
Then any point on $$(1)$$ is $$\left( 2r+1,3r-2,-6r+3 \right) $$.
If this point lies on the plane $$x-y+z=5$$, then
$$\displaystyle \left( 2r+1 \right) -\left( 3r-2 \right) +\left( -6r+3 \right) =5\Rightarrow -7r+6=5\Rightarrow r=\dfrac { 1 }{ 7 } $$
Hence, the point is $$\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 } \right) $$
Distance between $$\left( 1,-2,3 \right) $$ and $$\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 } \right) $$
$$\displaystyle =\sqrt { \left( \dfrac { 4 }{ 49 } +\dfrac { 9 }{ 49 } +\dfrac { 36 }{ 49 } \right) } =\sqrt { \dfrac { 49 }{ 49 } } =1$$
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