JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Three Dimensional Geometry

Any plane through the line of intersection is $P_1+\lambda P_2=0$, that is $(1+2\lambda )x+(1+3\lambda )y+(1+4\lambda )z=1+5\lambda$. For this to be perpendicular to $x-y+z=0$ the normals must be perpendicular, so $(1+2\lambda )-(1+3\lambda )+(1+4\lambda )=0$, giving $1+3\lambda =0$ and $\lambda =-\frac{1}{3}$. Substituting back, the plane becomes $x-z+2=0$, which in vector form is $\vec{r}\cdot (\hat{i}-\hat{k})+2=0$.