Given plane is $2x+y-3z=6$.

Since, $\left(\alpha ,\beta ,\gamma \right)$ is the image point of $\left(2,3,5\right)$ in the given plane, so

$\frac{\alpha -2}{2}=\frac{{\displaystyle \beta -3}}{{\displaystyle 1}}=\frac{{\displaystyle \gamma -5}}{{\displaystyle -3}}=-\frac{2\left(4+3-15-6\right)}{4+1+9}$

$\Rightarrow \frac{\alpha -2}{2}=\frac{{\displaystyle \beta -3}}{{\displaystyle 1}}=\frac{{\displaystyle \gamma -5}}{{\displaystyle -3}}=2$

$\Rightarrow \frac{\alpha -2}{2}=2; \frac{{\displaystyle \beta -3}}{{\displaystyle 1}}=2; \frac{{\displaystyle \gamma -5}}{{\displaystyle -3}}=2$

$\therefore \alpha =6, \beta =5, \gamma =-1$

Hence, $\alpha +\beta +\gamma =6+5-1=10$

Hence this is the required option.