Question
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.
Solution
The equation of any plane through the line of intersection of the planes
x + y + z – 1 = 0 and 2x + 3 y + 4 z – 5 = 0 is
(x + y + z – 1) + k (2 x + 3 y + 4 z – 5) = 0 ...(1)
or (2 k + 1) x + (3 k + 1) y + (4 k + 1) z – (1 + 5 k) = 0
∴ this plane is perpendicular to the plane x – y + z = 0
∴ (2 k + 1) (1) + (3 k + 1) (– 1) + (4 k + 1) (1) = 0
∴ 2 k + 1 – 3 k – 1 + 4 k + 1= 0
Putting 
(x + y + z - 1) - 
or 3 (x + z –1) – (2 x + 3 y + 4 z – 5) = 0
or 3 x + 3 y + 3 z – 3 – 2 x – 3 y – 4 z + 5= 0
or x – z + 2 = 0
which is required equation of plane.
