Question
Find the equation of the plane passing through the line of intersection of planes x + y – 2 z + 3 = 0 and 3 x – y – 2 z – 4 = 0 and perpendicular to the plane 2x + 3 y – z + 1 = 0.
Solution
Any plane through the line of intersection of planes
x + y – 2 z + 3 = 0 and 3 x – y – 2 z – 4 = 0 is
(x + y – 2 z + 3) + k (3 x – y – 2 z – 4) = 0 ...(1)
i.e. (3 k + 1) x + (– k + 1) y + (– 2 k – 2) z + (– 4 k + 3) = 0
Direction ratios of its normal are 3 k + 1, – k + 1, – 2 k – 2.
Again consider the plane
2 x + 3 y – z + 1 = 0 ...(2)
Direction ratios of its normal are 2, 3, – 1.
Since plane (1) is perpendicular to plane (2)
∴ (2) (3 k + 1) + (3) (– k + 1) + (– 1) (–2 k –2) = 0
∴ 6 k + 2 – 3 k + 3 + 2 k + 2 = 0
Putting 
or 5 (x + y – 2 z + 3) – 7 (3 x – y – 2 z – 4) = 0
or 5x + 5 y – 10 z + 15 – 21 x + 7 y + 14 z + 28 = 0
or – 16 x + 12 y + 4 z + 43 = 0
or 16 x – 12 y – 4z – 43 = 0
which is required equation of plane
