JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Three Dimensional Geometry

The plane through $(-2,3,5)$ has equation $a(x+2)+b(y-3)+c(z-5)=0$. Since it is perpendicular to both given planes, the normal vector $(a,b,c)$ is perpendicular to both $(2,4,5)$ and $(3,-2,3)$. This gives $2a+4b+5c=0$ and $3a-2b+3c=0$. Solving using cross product: $\frac{a}{22}=\frac{b}{9}=\frac{c}{-16}$. So the plane is $22(x+2)+9(y-3)-16(z-5)=0$, which simplifies to $22x+9y-16z+97=0$. Comparing with $\alpha x+\beta y+\gamma z+97=0$, we get $\alpha =22,\beta =9,\gamma =-16$. Therefore $\alpha +\beta +\gamma =22+9-16=15$.