Question

Equation of plane which passes through (1,-3,-2) and perpendicular to planes x+2y+2z=5 & 3x+3y+2z=8

Answer 1

Given that the plane passes through (1,-3,-2)

Since the plane is perpendigular to planes x+2y+2z=5 & 3x+3y+2z=8

we have our plane normal perpendicular to normals of both the planes.

The normal to both the planes can be found out using cross product.

Hence the required plane would be

$\left[\begin{array}{ccc}x-1&y+3&z+2\\1&2&2\\3&2&2\end{array}\right] =0$

Simplify to get

2x-4y+3z =8

Answer is

2x-4y+3z =8