Let point $\mathrm{S}(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }) \mathrm{be} \mathrm{an} \mathrm{image} \mathrm{of} \mathrm{Q}(1,3,4)$then$\frac{\alpha -1}{2}=\frac{\beta -3}{-1}=\frac{\gamma -4}{1}=\frac{-2(2-3+4+3)}{4+1+1}=-2$ $\Rightarrow S(\alpha ,\beta ,\gamma )=(-3,5,2)$ and

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

put the points of R in equation of plane anf find the value
$\Rightarrow 6-5+\gamma +3=0\Rightarrow \gamma =-4$

$\Rightarrow R(3,5,-4)\mathrm{\&}S(-3,5,2)$

using distance formula find

distance of $SR=\sqrt{36+0+36}\Rightarrow (SR{)}^2=72$