Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let $a,b,c$ be the direction cosines of the required plane. It contains the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and passes through the point $(0,7,-7)$ $$\begin{array}{rl}& \therefore \mathrm{a}(\mathrm{x}+1)+\mathrm{b}(\mathrm{y}-3)+\mathrm{c}(\mathrm{z}+2)=0\dots (1)\\ & \therefore \mathrm{a}(0+1)+\mathrm{b}(7-3)+\mathrm{c}(-7+2)=0\Rightarrow \mathrm{a}+4\mathrm{b}-5\mathrm{c}=0.\end{array}$$ $$\text{ Also }-3\mathrm{a}+2\mathrm{b}+\mathrm{c}=0$$ From (2) and (3), we write $$\frac{\mathrm{a}}{\left|\begin{array}{cc}4& -5\\ 2& 1\end{array}\right|}=\frac{-b}{\left|\begin{array}{cc}1& -5\\ -3& 1\end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{cc}1& 4\\ -3& 2\end{array}\right|}$$ $$\therefore \frac{\mathrm{a}}{14}=\frac{\mathrm{b}}{14}=\frac{\mathrm{c}}{14}\Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c}=1$$ Hence eq. (1) becomes $$x+1+y-3+z+2=0\Rightarrow x+y+z=0$$