Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Equation of plane passing through $(-2,2,2)$ is $a(x+2)+b$ $$(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-2)=0$$ Since this plane also passes through $(2,-2,-2)$, we get $$4\mathrm{a}-4\mathrm{b}-4\mathrm{c}=0\Rightarrow \mathrm{a}-\mathrm{b}-\mathrm{c}=0$$ Normal of the plane is parallel to $$9x-13y-3z=0\Rightarrow 9a-13b-3c=0$$ Solving (1) and (2), we write $$\frac{\mathrm{a}}{\left|\begin{array}{cc}-1& -1\\ -13& -3\end{array}\right|}=\frac{\mathrm{b}}{\left|\begin{array}{cc}1& -1\\ 0& -3\end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{cc}1& -1\\ 9& -13\end{array}\right|}$$ $$\therefore \frac{\mathrm{a}}{-10}=\frac{-\mathrm{b}}{6}=\frac{\mathrm{c}}{-4}\Rightarrow \frac{\mathrm{a}}{5}=\frac{\mathrm{b}}{3}=\frac{\mathrm{c}}{2}$$ Hence equation of required plane is $$5(x+2)+3(y-2)+2(z-2)=0\text{ i.e. }5x+3y+2z=0$$