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

2x + 3y–z + 1 + λ(x + y–2z + 3)=0

(2 + λ)x + (3 + λ)y + (-1-2λ)z + 1 + 3λ=0 (2)

Now as the plane 3x-y-2z-4=0 is perpendicular the given plane,

For θ=90°, cos90°=0

(3)

On comparing with standard equations in Cartesian form,

Putting values in equation (3), we have

(2 + λ).3 + (3 + λ).(-1) + (-1-2λ).(-2)=0

6 + 3λ-3-λ + 2 + 4λ=0

5 + 6λ=0

Putting in equation(2)

7x + 13y + 4z-9=0

7x + 13y + 4z=9

So, required equation of plane is 7x + 13y + 4z=9.