JEE Advanced 2022 · Paper 1, Mathematics question 11

The common edge. $\vec n_1\times\vec n_2=(10,15,12)\times(-2,5,4)=(0,-64,80)\parallel(0,-4,5)$. The point $(0,4,0)$ lies on both planes, so the planes meet in
$$L:\ x=0,\ 5y+4z=20.$$

Which lines can be edges. The faces on $P_1$ and $P_2$ share $L$ as an edge.

• Every other edge of those two faces lies in $P_1$ or in $P_2$.
• The sixth edge joins the two remaining vertices (one on each plane, off $L$). It must be skew to $L$; if it met $L$, all four vertices would be coplanar.

(A) $(1,1,1)+t(0,0,1)$ has $x=1$, so it never meets $L$, and it is not parallel to $L$. It cuts $P_1$ and $P_2$ at different points, so it can be the sixth edge. ✓

(B) passes through $(6,0,0)\in P_1$ with direction $(-5,2,3)$. It reaches $x=0$ at $(0,2.4,3.6)$, where $5y+4z=26.4\ne20$. So it is skew to $L$ and can be the sixth edge. ✓

(C) passes through $(0,4,0)\in L$ and does not lie in either plane ($(-2,5,4)$ is normal to $P_2$). An edge through a point of $L$ would make the vertices coplanar. ✗

(D) passes through $(0,4,0)$ with direction $(1,-2,3)$, and $(1,-2,3)\cdot(-2,5,4)=0$. So the line lies in $P_2$ and can be an edge of the face on $P_2$. ✓