JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Three Dimensional Geometry

The normals are ${\vec{n}}_1=(1,-2,-2)$ with $|{\vec{n}}_1|=3$ and ${\vec{n}}_2=(2,-3,-6)$ with $|{\vec{n}}_2|=7$. The bisector planes satisfy $\frac{x-2y-2z+1}{3}=\pm \frac{2x-3y-6z+1}{7}$. Both constant terms are positive and ${\vec{n}}_1\cdot {\vec{n}}_2=20>0$, so the acute-angle bisector corresponds to the negative sign: The point $\left(-2,0,-\frac{1}{2}\right)$ satisfies $13(-2)-23(0)-32\left(-\frac{1}{2}\right)+10=0$, so it lies on $P$.