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

Equation of plane through the line of intersection of planes in Cartesian form is

(1)

For the standard equation of planes,

So, putting in equation (1), we have

x + y + z-6 + λ(2x + 2y + 4z + 5)=0

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

Now plane passes through (1,1,1) then it must satisfy the plane equation,

(1 + 2λ).1 + (1 + 2λ).1 + (1 + 4λ).1-6 + 5λ=0

1 + 2λ + 1 + 2λ + 1 + 4λ-6 + 5λ=0

3 + 8λ-6 + 5λ=0

13λ=3

Putting in equation (2)

19x + 19y + 25z-63=0

So, the required equation of plane is 19x + 19y + 25z=63.