Question

The equation of the plane through the line x + y + z + 3 = 0 = 2x − y + 3z + 1 and parallel to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ is
(a) x − 5y + 3z = 7
(b) x − 5y + 3z = −7
(c) x + 5y + 3z = 7
(d) x + 5y + 3z = −7

Answer 1

(a) x − 5y + 3z = 7

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