Question

Find the equation of the plane passing through the line of intersection of the planes 2x − 7y + 4z − 3 = 0, 3x − 5y + 4z + 11 = 0 and the point (−2, 1, 3).

Answer 1

$$\begin{gathered} \text{The equation of the plane passing through the line of intersection of the given planes is} \\ 2x-7y+4z-3+\lambda \left(3x-5y+4z+11\right)=0...\left(1\right) \\ \text{This passes through (-2, 1, 3). So,} \\ -4-7+12-3+\lambda \left(-6-5+12+11\right)=0 \\ \Rightarrow -2+12\lambda =0 \\ \Rightarrow \lambda =\frac{1}{6} \\ \text{Substituting this in (1), we get} \\ 2x-7y+4z-3+\frac{1}{6}\left(3x-5y+4z+11\right)=0 \\ \Rightarrow 12x-42y+24z-18+3x-5y+4z+11=0 \\ \Rightarrow 15x-47y+28z=7 \end{gathered}$$