Question

Find the equations of the planes parallel to the plane $x+2y-2z+8=0$ which are at distance of $2$ units from the point $(2,1,1)$.

Answer 1

Equation of parallel plane,

$\begin{array}{c}x+2y-2z+\lambda =0\\ Distancefrom\left(2,1,1\right)\\ \left|\frac{2+2\left(1\right)-2\left(1\right)+\lambda }{\sqrt{1^2+2^2+{\left(-2\right)}^2}}\right|=2\\ \left|\frac{2+\lambda }{\sqrt{9}}\right|=2\\ \lambda =\pm 6-2\\ \lambda =-8,4\\ Hence,theequationoftheplanex+2y-2z-8=0orx+2y-2z+4=0\end{array}$.