Let S be the mirror image of the point

Since $\mathrm{R}(3,5, \gamma)$ lies on the plane $2 \mathrm{x}-\mathrm{y}+\mathrm{z}+3=0$

Therefore, $6-5+\gamma+3=0$

$\Rightarrow \gamma=-4$

Now,

dr's of line QS

are $2,-1,1$

equation of line $\mathrm{QS}$ is

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\lambda($ say $)$

$\Rightarrow \mathrm{F}(2 \lambda+1,-\lambda+3, \lambda+4)$

F lies in the plane

$\Rightarrow 2(2 \lambda+1)-(-\lambda+3)+(\lambda+4)+3=0$

$\Rightarrow 4 \lambda+2+\lambda-3+\lambda+7=0$

$\Rightarrow 6 \lambda+6=0 \Rightarrow \lambda=-1$

$\Rightarrow F(-1,4,3)$

Since, $\mathrm{F}$ is mid-point of $\mathrm{QS}$.

Therefore, co-ordinated of $S$ are $(-3,5,2)$.

So, SR $=\sqrt{36+0+36}=\sqrt{72}$

$\mathrm{SR}^{2}=72$