Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of planes parallel to the plane $x-2y+2z+4=0$ is $$\mathrm{x}-2\mathrm{y}+2\mathrm{z}+\lambda =0$$ The required planes are at a distance of one unit from $(1,2,3)$ $$\begin{array}{rl}& \left|\frac{{\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{cz}{z}_1+\mathrm{d}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}\right|=1\\ & \left|\frac{1(1)+(-2)(2)+2(3)+\lambda }{\sqrt{1+4+4}}\right|=1\Rightarrow \left|\frac{1-4+6+\lambda }{3}\right|=1\Rightarrow 3+\lambda =\pm 3\\ & \therefore 3+\lambda =3\text{ or }3+\lambda =-3\\ & \lambda =0\text{ or }\lambda =-6\end{array}$$ The equation of planes are $x-2y+2z=0$ and $x-2y+2z-6=0$