A plane passes through the point $(1, 1, 1)$ and is perpendicular to each of the planes $3x − 2y + 3z + 6 = 0$ and $6x − 2y − 3z − 6 = 0$

Question

I need help please on this question A plane passes through the point $(1, 1, 1)$ and is perpendicular to each of the planes

$3x − 2y + 3z + 6 = 0$ and $6x − 2y − 3z − 6 = 0$. Find its equation. The problem is I don't have an idea of the concept. All I know is that the normal of first equation is $(3,-2,3)$ and that of the second is $(6,-2,-3).$

Answer 1

Here’s a hint: the normal of the third plane is perpendicular to both normals of the two given planes. Use the cross product.

Answer 2

So if $T(x,y,z)$ is in this plane and $A(1,1,1)$ then $$\vec{AT}=(x-1,y-1,z-1) = m(3,-2,3)+n(6,-2,-3)$$ for some scalars $m,n$. Eliminate the scalars and you are done.