Question
Find the direction ratios of the normal to the plane passing through the point (2,1, 3) and the line of intersection of the planes x + 2 y + z = 3 and 2 x – y – z = 5.
Solution
The equation of any plane through the intersection of planes
x + 2y + z – 3 = 0 and 2x – y – z – 5 = 0 is
(x + 2y + z – 3) + k (2x – y – z – 5) = 0 ...(1)
∴ it passes through (2, 1, 3)
∴ (2 + 2 + 3 – 3) + k (4 – 1– 3 – 5) = 0
Putting 
or 5 (x + 2 y + z – 3) + 4 (2 x – y – z – 3) = 0
or 5 x + 10 y + 5 z – 15 + 8 x – 4 y – 4 z – 12 = 0
or 13 x + 6 y + z – 27 = 0
which is equation of plane.
Direction ratios of normal to the plane are 13, 6, 1.
