Question

Find the equation of the plane through the intersection of the planes 3x − 4y + 5z = 10 and 2x + 2y − 3z = 4 and parallel to the line x = 2y = 3z.

Answer 1

$$\begin{gathered} \text{The equation of the plane passing through the intersection of the given planes is} \\ \left(3x-4y+5z-10\right)+\lambda \left(2x+2y-3z-4\right)=0 \\ \Rightarrow \left(3+2\lambda \right)x+\left(-4+2\lambda \right)y+\left(5-3\lambda \right)z-10-4\lambda =0...\left(1\right) \\ \text{The given line is} \\ x=2y=3z \\ \text{Dividing this equation by 6, we get} \\ \frac{x}{6}=\frac{y}{3}=\frac{z}{2} \\ \text{The direction ratios of this line are proportional to 6, 3, 2.} \\ \text{So, the normal to the plane is perpendicular to}\mathrm{the\; line\; whose\; direction\; ratios\; are\; proportional\; to}\mathrm{6,\; 3,\; 2}. \\ \Rightarrow \left(3+2\lambda \right)6+\left(-4+2\lambda \right)3+\left(5-3\lambda \right)2=0 \\ \Rightarrow 18+12\lambda -12+6\lambda +10-6\lambda =0 \\ \Rightarrow 12\lambda +16=0 \\ \Rightarrow \lambda =\left(\frac{-4}{3}\right) \\ \text{Substituting this in (1), we get} \\ \left(3+2\left(\frac{-4}{3}\right)\right)x+\left(-4+2\left(\frac{-4}{3}\right)\right)y+\left(5-3\left(\frac{-4}{3}\right)\right)z-10-4\left(\frac{-4}{3}\right)=0 \\ \Rightarrow x-20y+27z=14 \end{gathered}$$