Question

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.

Answer 1

$$\begin{gathered} \text{The equation of the plane passing through the line of intersection of the given planes is} \\ 2x+3y-z+1+\lambda \left(x+y-2z+3\right)=0 \\ \left(2+\lambda \right)x+\left(3+\lambda \right)y+\left(-1-2\lambda \right)z+1+3\lambda =0...\left(1\right) \\ \text{This plane is perpendicular to 3}x-y-2z-4=0.\text{So,} \\ 3\left(2+\lambda \right)-\left(3+\lambda \right)-2\left(-1-2\lambda \right)=0\text{(Because}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\text{)} \\ \Rightarrow 6+3\lambda -3-\lambda +2+4\lambda =0 \\ \Rightarrow 6\lambda +5=0\mathit{} \\ \Rightarrow \lambda =\frac{-5}{6} \\ \text{Substituting this in (1), we get} \\ \left(2-\frac{5}{6} \\ \right)x+\left(3-\frac{5}{6} \\ \right)y+\left(-1-2\left(\frac{-5}{6} \\ \right)\right)z+1+3\left(\frac{-5}{6} \\ \right)=0 \\ \Rightarrow 7X+13Y+4z-9=0 \end{gathered}$$