The linear equation passing through point (1, - 1, - 2) and perpendicular to plane 2x-2y + 3Z = 0

The linear equation passing through point (1, - 1, - 2) and perpendicular to plane 2x-2y + 3Z = 0

If the normal vector of plane 2x-2y + 3Z = 0 is (2, - 2,3), take any point (x, y, z) in the line, then the linear equation is 2x-2y + 3Z = a, and bring the point (1, - 1, - 2) into a = - 2, so the linear equation is 2x-2y + 3Z + 2 = 0