If the mirror image

Point $Q$ is image of point $P$ w.r.to plane, $M$ is mid point of $P$ and $Q$, lies in plane

$\mathrm{M}\left(\frac{1+\alpha}{2}, \frac{3+\beta}{2}, \frac{5+\gamma}{2}\right)$

$4 x-5 y+2 z=8$

$4\left(\frac{1+\alpha}{2}\right)-5\left(\frac{3+\beta}{2}\right)+2\left(\frac{5+\gamma}{2}\right)=8$..(1)

Also PQ perpendicualr to the plane $\Rightarrow \overrightarrow{\mathrm{PQ}} \| \overrightarrow{\mathrm{n}}$

$\frac{\alpha-1}{4}=\frac{\beta-3}{-5}=\frac{\gamma-5}{2}=\mathrm{k} \quad$ (let)

$\left.\begin{array}{l}\alpha=1+4 \mathrm{k} \\ \beta=3-5 \mathrm{k} \\ \gamma=5+2 \mathrm{k}\end{array}\right\}$..(2)

use $(2)$ in $(1)$

$2(1+4 \mathrm{k})-5\left(\frac{6-5 \mathrm{k}}{2}\right)+(10+2 \mathrm{k})=8$

$\mathrm{k}=\frac{2}{5}$

from (2) $\alpha=\frac{13}{5}, \beta=1, \gamma=\frac{29}{5}$

$5(\alpha+\beta+\gamma)=13+5+29=47$