Find the equation of the plane passing through the intersection of the planes `2x+3y-z+1=0` and `x+y-2z+3=0` and perpendicular to the plane `3x-y-2z-4=0.`

Explanation

Given equation of planes are
`2x+3y−z+6=0` and `x+y−2z+3=0`
As we know that

The equation of plane passing through the line of intersection of the planes
`a_1x+b_1y+c_1z+d_1=0` and `a_2x+b_2y+c_2z+d_2=0` is
`a_1x+b_1y+c_1z+d_1+lambda(a_2x+b_2y+c_2z+d_2)=0`
So the required equation of plane is `2x+3y−z+1+λ(x+y−2z+3)=0`
`=>(2+λ)x+(3+λ)y+(−1−2λ)z+1+3λ=0⋯(1)`
This plane is perpendicular to `3x−y−2z−4=0`
`3(2+λ)−(3+λ)−2(1−2λ)=0`
`=>λ=− 5/6`
Substituting the `λ` value in `(1)`
`=>7x+13y+4z−9=0`
`∴ `Equation of required plane is `7x+13y+4z−9=0`