The plane containing the line $\frac{x-3}{2}= \frac{y+2}{-1}=\frac{z-1}{3}$ and also containing its projection on the plane $2x + 3y – z = 5$, contains which one of the following points? $(A) (2, 2, 0) (B) (–2, 2, 2) (C) (0, –2, 2) (D) (2, 0, –2)$
I think that plane would be $a(x-3)+b(y+2)+c(z-1)=0$, where $a,b,c$ are direction ratios of the normal of the plane. So, $2a-b+3c=0$. I don't understand how the projection of the line on another plane could lie on the first plane. Does that mean the two planes are intersecting? Do they have to perpendicular for this? Because, I guess this is what the solution is implying as it has written $2a+3b-c=0$.
See that since the plane contains the line and and it's projection onto the plane $2x+3y–z=5$, it means that the plane would contain the normal to the plane $2x+3y–z=5$ and yes, would be perpendicular to it.
To find the direction ratios of the plane , you just need to find the cross product of $(2,3,-1)$ and $(2,-1,3)$ as both the directions are parallel to the plane.