JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Three Dimensional Geometry

Every plane through the given line belongs to the family

that is $(1+\lambda )x+(2-\lambda )y+(3-\lambda )z+(1-6\lambda )=0$.
Perpendicularity to $-2x+y+z+8=0$ makes the two normals orthogonal:

Substituting and clearing the denominator, $P$ is $7x+5y+9z-14=0$.
Among the given points only $(0,1,1)$ satisfies it, since $0+5+9-14=0$, so $P$ passes through $(0,1,1)$.