Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ is $$a(x+1)+b(y-3)+c(z+2)=\mathrm{0...}(i)$$ where $-3a+2b+c=0$ ...(ii) This passes through $(0,7,-7)$. $\therefore$ $a+4b-5c=0$ ...(iii) From Eqs. (ii) and (iii), or $$\begin{array}{rl}\frac{a}{-14}& =\frac{b}{-14}=\frac{c}{-14}\\ \frac{a}{1}& =\frac{b}{1}=\frac{c}{1}\end{array}$$ Thus, the required plane is $x+y+z=0$