Three Dimensional Geometry Part 2

Equation of plane containing the given planes is
(x+2y+3z+1)+λ(x−y−z−6)=0
(1+λ)x+(2−λ)y+(3−λ)z+(1−6λ)=0
This plane is perpendicular to the plane −2x+y+z+8=0
So, −2(1+λ)+(2−λ)+(3−λ)=0
−2−2λ+2−λ+3−λ=0
λ=

3

4

So, equation of required plane is 7x+5y+9z=14
Now, (0,1,1) satisfies the above plane.