Question

Find the equation of the plane containing the line of intersection of the planes x + y + z − 6 = 0 and 2x + 3y + 4y + 5 = 0 and passing through the point (1, 1, 1)

Answer 1

The equation of a plane through the line of intersection of the given plane is

(x + y + z − 6) + λ(2x + 3y + 4z + 5) = 0   (1)

If Eq. (1) passes through (1, 1, 1), we have

−3 + 14λ = 0 ⇒ λ = 3/14

Putting λ = 3/14 in Eq. (1), we obtain the equation of the required plane as

(x + y + z − 6) + 3/14(2x + 3y + 4z + 5) = 0

⇒ 20x + 23y + 26z − 69 = 0