Question

Find the equation of the plane that is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z − 4 = 0, 2x + y − z + 5 = 0.

Answer 1

$$\begin{gathered} \text{The equation of the plane passing through the line of intersection of the given planes is} \\ x+2y+3z-4+\lambda \left(2x+y-z+5\right)=0 \\ \left(1+2\lambda \right)x+\left(2+\lambda \right)y+\left(3-\lambda \right)z-4+5\lambda =0...\left(1\right) \\ \text{This plane is perpendicular to 5}x+3y+6z+8=0.\text{So,} \\ 5\left(1+2\lambda \right)+3\left(2+\lambda \right)+6\left(3-\lambda \right)=0\text{(Because}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\text{)} \\ \Rightarrow 5+10\lambda +6+3\lambda +18-6\lambda =0 \\ \Rightarrow 7\lambda +29=0 \\ \Rightarrow \lambda =\frac{-29}{7} \\ \text{Substituting this in (1), we get} \\ \left(1+2\left(\frac{-29}{7}\right)\right)x+\left(2-\frac{29}{7}\right)y+\left(3+\frac{29}{7}\right)z-4+5\left(\frac{-29}{7}\right)=0 \\ \Rightarrow -51x-15y+50z-173=0 \\ \Rightarrow 51x+15y-50z+173=0 \end{gathered}$$