Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

The equation of the plane passing through the point (−1, 3, 2) is

a (x + 1) + b (y − 3) + c (z − 2) = 0 … (1)

where, a, b, c are the direction ratios of normal to the plane.

It is known that two planes, and , are perpendicular, if

Plane (1) is perpendicular to the plane, x + 2y + 3z = 5

Also, plane (1) is perpendicular to the plane, 3x + 3y + z = 0

From equations (2) and (3), we obtain

Substituting the values of a, b, and c in equation (1), we obtain

This is the required equation of the plane.