Let ( α, β, γ ) be the image of point P ( 2, 3, 5 ) in the plane 2x...

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

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

$⇒ \dfrac{\alpha - 2}{2} = \dfrac{\beta - 3}{1} = \dfrac{\gamma - 5}{- 3} = 2$

$⇒ \dfrac{\alpha - 2}{2} = 2; \dfrac{\beta - 3}{1} = 2; \dfrac{\gamma - 5}{- 3} = 2$

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

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

Hence this is the required option.