Question

Find the equation of the plane determined by the intersection of the lines $\frac{x+3}{3}=\frac{y}{-2}=\frac{z-7}{6}\mathrm{and}\frac{x+6}{1}=\frac{y+5}{-3}=\frac{z-1}{2}$

Answer 1

$$\begin{gathered} \text{The given equations of the lines are} \\ \frac{x+3}{3}=\frac{y}{-2}=\frac{z-7}{6}...\left(1\right) \\ \frac{x+6}{1}=\frac{y+5}{-3}=\frac{z-1}{2}...\left(2\right) \\ \text{Let the direction ratios of the plane be proportional to}a,b,c. \\ \text{Since the plane contains line (1), it should pass through (-3, 0, 7) and is parallel to the line (1).} \\ \text{Equation of the plane through (1) is} \\ a\left(x+3\right)+b\left(y\right)+c\left(z-7\right)=0...\left(3\right), \\ \text{where}3a-2b+6c=0...\left(4\right) \\ \text{Since the plane contains line (2), the plane is parallel to line (2) also.} \\ \Rightarrow a-3b+2c=0...\left(5\right) \\ \text{Solving (4) and (5) using cross-multiplication, we get} \\ \frac{a}{14}=\frac{b}{0}=\frac{c}{-7} \\ \text{Substituting}a,b\text{and}\mathit{\text{c}}\text{in (3), we get} \\ 14\left(x+3\right)+0\left(y\right)-7\left(z-7\right)=0 \\ \Rightarrow 2\left(x+3\right)+0\left(y\right)-1\left(z-7\right)=0 \\ \Rightarrow 2x-z+13=0 \end{gathered}$$