JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Three Dimensional Geometry

To find mirror image Q of P(1,0,1) with respect to plane $x+y+z=5$:

The perpendicular from P to the plane has direction ratios $(1,1,1)$.

$\frac{a-1}{1}=\frac{b-0}{1}=\frac{c-1}{1}=\frac{-2(1+0+1-5)}{1^2+1^2+1^2}=\frac{-2(-3)}{3}=2$

So $Q=(3,2,3)$.

Line L passes through $(1,-1,-1)$ parallel to $\overrightarrow{PQ}=(2,2,2)$ or direction $(1,1,1)$:

$\frac{x-1}{1}=\frac{y+1}{1}=\frac{z+1}{1}=\lambda$

Point on L: $(\lambda +1,\lambda -1,\lambda -1)$

This lies on plane $x+y+z=5$:

$(\lambda +1)+(\lambda -1)+(\lambda -1)=5$

$3\lambda -1=5\Rightarrow \lambda =2$

So $R=(3,1,1)$.

$Q{R}^2=(3-3{)}^2+(2-1{)}^2+(3-1{)}^2=0+1+4=5$.