JEE Main 2022 — Three Dimensional Geometry Question with Solution

Let a point on the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}=\lambda$ be

$M\left(2\lambda -1,3\lambda +3,-\lambda +1\right)$

and the DR's of the line are 

The DR's of line joining the points $\left(2\lambda -1,3\lambda +3,-\lambda +1\right)$ and $\left(a,4,2\right)$ will be

Now its perpendicular to the given line, so$\left(2\lambda -1-a\right)2+\left(3\lambda -1\right)3+\left(-\lambda -1\right)\left(-1\right)=0$

$\Rightarrow 4\lambda -2-2\mathrm{a}+9\lambda -3+\lambda +1=0$

$\Rightarrow 14\lambda -4-2\mathrm{a}=0$

$\Rightarrow 7\lambda -2-a=0\Rightarrow 7\lambda -2=a$

and,

${\left(2\lambda -1-a\right)}^2+{\left(3\lambda -1\right)}^2+{\left(\lambda +1\right)}^2={\left(2\sqrt{6}\right)}^2$

$\Rightarrow {\left(5\lambda -1\right)}^2+{\left(3\lambda -1\right)}^2+{\left(\lambda +1\right)}^2=24$

$\Rightarrow 35{\lambda }^2-14\lambda -21=0$

$\Rightarrow \left(\lambda -1\right)\left(35\lambda +21\right)=0$

For, $\lambda =1; a=5$

i.e. $M\equiv \left(1,6,0\right)$

As $Q\left({\alpha }_1,{\alpha }_2,{\alpha }_3\right)$ is the image of point $P$

$\begin{array}{ccc}\frac{{\alpha }_1+5}{2}=1;& \frac{{\alpha }_2+4}{2}=6;& \frac{{\alpha }_3+2}{2}=0\end{array}$

$\Rightarrow \begin{array}{ccc}{\alpha }_1=-3;& {\alpha }_2=8;& {\alpha }_3=-2\end{array}$

Hence, $a+{\alpha }_1+{\alpha }_2+{\alpha }_3=8$