Question

Find the equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point (0, 7, −7) and show that the line $\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}$ also lies in the same plane.

Answer 1

$$\begin{gathered} \text{Let the equation of the plane passing through (0, 7, -7) be} \\ a\left(x-0\right)+b\left(y-7\right)+c\left(z+7\right)=0...\left(1\right) \\ \text{The line}\frac{x+1}{-3}\text{=}\frac{y-3}{2}\text{=}\frac{z+2}{1}\text{passes through (-1, 3, -2) and its direction ratios are proportional to -3, 2, 1.} \\ \text{Since plane (1) contains this line, it must pass through the point (-1, 3, -2).} \\ \Rightarrow a\left(-1-0\right)+b\left(3-7\right)+c\left(-2+7\right)=0 \\ \Rightarrow -a-4b+5c=0 \\ \Rightarrow a+4b-5c=0...\left(2\right) \\ \mathrm{Since}\mathrm{plane\; (1)\; contains\; this\; line,}\text{it must be parallel to the line.} \\ \Rightarrow -3a+2b+c=0...\left(3\right) \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x-0& y-7& z+7\\ 1& 4& -5\\ -3& 2& 1\end{array}\right|=0 \\ \Rightarrow 14x+14\left(y-7\right)+14\left(z+7\right)=0 \\ \Rightarrow 14x+14y+14z=0 \\ \Rightarrow x+y+z=0 \end{gathered}$$