JEE Main 2023 (Online) 11th April Evening ShiftQuestion 1
Let $P$ be the plane passing through the points $(5,3,0)$, $(13,3,-2)$ and $(1,6,2)$. For $\alpha\in\mathbb{N}$, if the distances of the points $A(3,4,\alpha)$ and $B(2,\alpha,a)$ from the plane $P$ are $2$ and $3$ respectively, then the positive value of $a$ is
3D Geometry
Solution
Plane: $3x-4y+12z=3$ (norm $13$). $A$: $|12\alpha-10|=26$ gives $\alpha=3$. $B(2,3,a)$: $|12a-9|=39$ gives $a=4$ (positive).