Question
Find the equation of the plane passing through the points (3, 2, 2) and (1, 0, –1) parallel to the line 
Solution
The equation of plane through (3, 2, 2) is
A (x – 3) + B (y – 2) + C (z – 2) = 0 ...(1)
Since it passes through (1, 0, –1)
∴ A (1 – 3) + B (0 – 2) + C (–1 – 2) = 0
or –2A – 2B – 3C = 0
∴ 2A + 2B + 3C = 0 .,.(2)
The equation of line is
Its direction ratios are 2, –2, 3
Since the line is parallel to plane (1) whose normal has direction ratios A, B, C.
∴ normal to plane (1) is perpendicular to line
∴ 2A – 2B + 3C = 0 ...(3)
From (2) and (3), we get,
Putting these values of A, B, C in (1), we get,
3k (x – 3) + 0 (y – 2) – 2k (z –2) = 0
or 3 (x – 3) – 2 (z – 2 ) = 0
or 3x – 9 –2z + 4 = 0 or 3x –2z – 5 = 0
