JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Three Dimensional Geometry

Two points of the plane are $(0,-1,2)$ and $(-1,2,1)$, giving an in-plane direction $\overrightarrow{v_1}=(-1,3,-1)$. The plane is parallel to the line through $(5,1,-7)$ and $(1,-1,-1)$, giving a second direction $\overrightarrow{v_2}=(-4,-2,6)$. The normal is $\vec{n}=\overrightarrow{v_1}\times \overrightarrow{v_2}=(16,10,14)=2(8,5,7)$, so the plane is $8x+5y+7z=d$. Using $(0,-1,2)$, $-5+14=d$, hence $d=9$ and the plane is $8x+5y+7z=9$. Among the given points, $8(-2)+5(5)+7(0)=-16+25=9$, so the plane passes through $(-2,5,0)$.