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

Let parallel vector of $L=\overrightarrow{b}$

mirror image of Q on given plane $x+y+z=5$

$$\begin{gathered} \frac{a-1}{1}=\frac{b-0}{1}=\frac{c-1}{1}=\frac{-2(2-5)}{3} \\ \mathrm{a}=3, \mathrm{b}=2,\mathrm{c}=3 \\ Q\equiv (3,2,3) \\ \because \overrightarrow{b}\Vert \overrightarrow{PQ} \\ \mathrm{so}, \overrightarrow{\mathrm{b}}=(1,1,1) \end{gathered}$$

Equation of line

$$\begin{gathered} \mathrm{L}:\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+1}{1}=\frac{\mathrm{z}+1}{1} \\ \text{ Let point }\mathrm{R},(\lambda +1,\lambda -1,\lambda -1) \\ \text{ lying on plane }x+y+z=5\text{, } \\ \text{ so, }3\lambda -1=5 \\ \Rightarrow \lambda =2 \end{gathered}$$

$$\begin{gathered} \text{ Point }\mathrm{R}\text{ is }(3,1,1) \\ {\mathrm{QR}}^2=5 \end{gathered}$$