Question

If the equation of the plane through the intersection of the planes x + 2y + 3z − 4 = 0 and 2x + y − z + 5 = 0 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is ax + by + cz + 173 = 0, then b − 9 (a + c) is equal to _______.

Answer 1

The required plane is of the form

whose normal is 

This plane is perpendicular to the plane 5x + 3y + 6z + 8 = 0. So we have

Therefore, the required plane is

(x + 2y + 3z - 4) - 29/7(2x + y - z + 5) = 0

51x + 15y - 50z + 173 = 0

comparing this with ax + by + cz + 173 = 0 we get a = 51, b = 15, c = 50. so that 

b − 9 (a + c) = 15 − 9 = 6