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Q.
Let $Q$ be the mirror image of the point $P(1,0,1)$ with respect to the plane $S : x + y + z =5$. If a line $L$ passing through $(1,-1,-1)$, parallel to the line $PQ$ meets the plane $S$ at $R$, then $QR^{2}$ is equal to:
Let parallel vector of $L=\vec{b}$
mirror image of $Q$ on given plane $x+y+z=5$
$\frac{a-1}{1}=\frac{b-0}{1}=\frac{c-1}{1}=\frac{-2(2-5)}{3} $
$a=3, b=2, c=3 $
$Q \equiv(3,2,3)$
$\because \vec{b}|| \overrightarrow{P Q}$
so,$ \vec{b}=(1,1,1)$
Equation of line
$L : \frac{ x -1}{1}=\frac{ y +1}{1}=\frac{ z +1}{1}$
Let point $R ,(\lambda+1, \lambda-1, \lambda-1)$
lying on plane $x + y + z =5$,
so, $ 3 \lambda-1=5 $
$\Rightarrow \lambda=2$
Point $R$ is $(3,1,1)$
$QR ^{2}=5 $
