JEE Main 2023 (Online) 8th April Evening ShiftQuestion 1
Let $P$ be the plane passing through the line $\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$ and the point $(2,4,-3)$. If the image of the point $(-1,3,4)$ in the plane $P$ is $(\alpha,\beta,\gamma)$ then $\alpha+\beta+\gamma$ is equal to
3D Geometry
Solution
Normal $=(1,-3,7)\times(1,2,2)\parallel(-4,1,1)$, plane $-4x+y+z+7=0$. At $(-1,3,4)$ the expression is $18=|n|^2$, so image $=(-1,3,4)-2(-4,1,1)=(7,1,2)$, sum $10$.