Question

Find the equation of the plane through the intersection of the planes 3x − y + 2z = 4 and x + y + z = 2 and the point (2, 2, 1).

Answer 1

$$\begin{gathered} \text{The equation of the plane passing through the line of intersection of the given planes is} \\ 3x-y+2z-4+\lambda \left(x+y+z-2\right)=0...\left(1\right) \\ \text{This passes through (2, 2, 1). So,} \\ 6-2+2-4+\lambda \left(2+2+1-2\right)=0 \\ \Rightarrow 2+3\lambda =0 \\ \Rightarrow \lambda =\frac{-2}{3} \\ \text{Substituting this in (1), we get} \\ 3x-y+2z-4-\frac{2}{3}\left(x+y+z-2\right)=0 \\ \Rightarrow 9x-3y+6z-12-2x-2y-2z+4=0 \\ \Rightarrow 7x-5y+4z=8 \end{gathered}$$