Let $Q$ be the mirror image of the point $P\left(1,0,1\right)$ with respect to the plane $S:x+y+z=5$. If a line $L$ passing through $\left(1,-1,-1\right)$, parallel to the line $PQ$ meets the plane $S$ at $R$, then $Q{R}^2$ is equal to

Let parallel vector of $L=$

Let $Q\left(a,b,c\right)$

As $Q$ is the mirror image on given plane $x+y+z=5$

so $\frac{a-1}{1}=\frac{b-0}{1}=\frac{c-1}{1}=\frac{-2\left(2-5\right)}{3}$

$a=3,b=2,c=3$

i.e. $Q\equiv \left(3,2,3\right)$

Let the vector parallel to $L$ be $\overrightarrow{b}$

So, $\overrightarrow{b}=\left(1,1,1\right)$

Equation of line $L:\frac{x-1}{1}=\frac{y+1}{1}=\frac{z+1}{1}$

Let point $R\left(\lambda +1,\lambda -1,\lambda -1\right)$ is lying on plane $x+y+z=5$, so, $3\lambda -1=5$

$\Rightarrow \lambda =2$
So, point $R$ is $\left(3,1,1\right)$

Hence, $Q{R}^2=5$