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

The equation of plane that passes through the point $(-2,3,5)$ is <br/><br/>$a(x+2)+b(y-3)+c(z-5)=0$ ..........(i) <br/><br/>The plane is perpendicular to <br/><br/>$$ \begin{array}{ll} 2 x+4 y+5 z =8 \text { and } 3 x-2 y+3 z=5 \\\\ \end{array} $$ <br/><br/>$$ \begin{array}{ll} \therefore 2 a+4 b+5 c=0 ..........(ii)\\\\ \text { and } 3 a-2 b+3 c=0 ..........(iii) \end{array} $$ <br/><br/>$$ \begin{aligned} & \therefore \quad \frac{\mathrm{a}}{\left|\begin{array}{cc} 4 & 5 \\ -2 & 3 \end{array}\right|}=\frac{-\mathrm{b}}{\left|\begin{array}{ll} 2 & 5 \\ 3 & 3 \end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{cc} 2 & 4 \\ 3 & -2 \end{array}\right|} \\\\ & \Rightarrow \frac{\mathrm{a}}{22}=\frac{\mathrm{b}}{9}=\frac{\mathrm{c}}{-16} \end{aligned} $$ <br/><br/>$\therefore$ From Equation (i) equation of plane <br/><br/>$$ \begin{aligned} &22(x+2)+9(y-3)-16(z-5)=0 \\\\ &\Rightarrow 22 x+9 y-16 z+97=0 \end{aligned} $$ <br/><br/>Here, $\alpha=22, \beta=9, \gamma=-16$ <br/><br/>$\therefore \alpha+\beta+\gamma=22+9-16=15$