If the mirror image of the point $\left(1,3,5\right)$ with respect to the plane $4x-5y+2z=8$ is $\left(\alpha ,\beta ,\gamma \right)$, then $5\left(\alpha +\beta +\gamma \right)$ equals :

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

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

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

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

Also $PQ$ perpendicualr to the plane

$\Rightarrow \overrightarrow{PQ}\Vert \overrightarrow{n}$

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

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

use $\left(2\right)$ in $\left(1\right)$

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

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

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

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