If the equation of the plane that contains the point (-2, 3, 5) and is perpendicular to each of the planes $2\mathrm{x}+4\mathrm{y}+5\mathrm{z}=8$ and $3\mathrm{x}-2\mathrm{y}+3\mathrm{z}=5$ is $\mathrm{\alpha x}+\mathrm{\beta y}+\mathrm{\gamma z}+97=0$ then $\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=$

Point (-2, 3, 5) lies on $\mathrm{\alpha x}+\mathrm{\beta y}+\mathrm{\gamma z}+97=0$
$\Rightarrow -2\mathrm{\alpha }+3\mathrm{\beta }+5\mathrm{\gamma }+97=0 ...\left(1\right)$
$2\mathrm{x}+4\mathrm{y}+5\mathrm{z}=8$ is perpendicular to given plane
$\Rightarrow 2\mathrm{\alpha }+4\mathrm{\beta }+5\mathrm{\gamma }=0 ...\left(2\right)$
$3\mathrm{x}-2\mathrm{y}+3\mathrm{z}=5$ is also perpendicular to given plane
$\Rightarrow 3\mathrm{\alpha }-2\mathrm{\beta }+3\mathrm{\gamma }=0 ...\left(3\right)$
Solving (1), (2), (3)
$$\begin{gathered} \mathrm{\alpha }=22,\mathrm{\beta }=9,\mathrm{\gamma }=-16 \\ \therefore \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=22+9-16=15 \end{gathered}$$