Question
Find the equation of the plane passing through the points (1, 2, 3) and (0. –1, 0) and parallel to the line 
Solution
The equation of any plane through (1, 2, 3) is
A(x – 1)+B(y – 2) + C(z – 3) = 0 ...(1)
∵ it passes through (0, – 1, 0)
∴ A (0 – 1) + B (– 1 – 2) + C (0 – 3) = 0
∴ – A – 3B – 3C = 0 ⇒ A + 3B + 3C = 0 ...(2)
Since plane (1) is parallel to the line 
∴ normal to the plane with direction ratios A, B, C is perpendicular to the line with direction ratios 2, 3, – 3.
∴ A(2) + B(3) + C(– 3) = 0 [∵ a1a2+ b1b2 + c1c2 = 0]
∴ 2A + 3B-3C = 0
Solving (2) and (3), we get,
Putting values of A, B, C in (1), we get,
6k(x – 1) + (–3 k)(y – 2) + k(z – 3) = 0 or 6 (x – 1) –3 (y – 2) + (z – 3) = 0
or 6x – 6 – 3y + 6 + z – 3 = 0
or 6x-3y + z=3
which is required equation of plane.
