JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Three Dimensional Geometry

The foot of perpendicular from $A(4,3,1)$ to $x-y+2z+3=0$ uses $t=-\frac{4-3+2+3}{6}=-1$, giving $N=(3,4,-1)$ and $AN=\sqrt{6}$. As $AN$ is perpendicular to the plane, triangle $ABN$ is right-angled at $N$, so area $=\frac{1}{2}AN\cdot BN=3\sqrt{2}$ gives $BN=2\sqrt{3}$ and $B{N}^2=12$. With $B(5,\alpha ,\beta )$ on the plane, $5-\alpha +2\beta +3=0$, so $\alpha =8+2\beta$; then $4+(\alpha -4{)}^2+(\beta +1{)}^2=12$ becomes $5{\beta }^2+18\beta +9=0$, whose integer root is $\beta =-3$, giving $\alpha =2$. Therefore ${\alpha }^2+{\beta }^2+\alpha \beta =4+9-6=7$.