Question
Find the equation of the plane through the points (2, 2, 1), (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z – 1 = 0.
Solution
The equation of any plane through (2, 2, 1) is
a (x – 2) + b (y – 2) + c (z – 1) = 0 ....(1)
∴ it passes through (9, 3, 6)
∴ a (9 – 2) + b (3 – 2) + c (6 – 1) = 0
∴ 7 a + b + 5 c = 0 ...(2)
Also plane (1) is perpendicular to the plane 2 x + 6 y + 6 z = 9
∴ a (2) + b (6) + c (6) = 0 [∴ a1 a2 + b1 b2 + c1 c2 = 0]
∴ 2 a + 6 b + 6 c = 0
⇒ a + 3 b + 3 c = 0 ....(3)
From (2) and (3), we get, 
∴ a = 3 k, b = 4 k, c = 5 k
Putting these values of a , b, c, in (1),
3 k (x – 2) + 4 k (y – 2) + (– 5 k) (z – 1) – 0
or 3 (x – 2) + 4 (y – 2) – 5 (z – 1) = 0
or 3 x – 6 + 4 y – 8 – 5 z + 5 = 0
or 3 x + 4 y – 5 z – 9 = 0
which is the required equation of plane.
