Question

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

Answer 1

$$\begin{gathered} \text{The equation of the plane parallel to the given plane is} \\ x+2y-2z+k=0...\left(1\right) \\ \text{It is given that plane (1) is at a distance of 2 units from (2, 1, 1).} \\ \Rightarrow \frac{\left|2+2-2+k\right|}{\sqrt{1^2+2^2+{\left(-2\right)}^2}}=2 \\ \Rightarrow \frac{\left|2+k\right|}{3}=2 \\ \Rightarrow \left|2+k\right|=6 \\ \Rightarrow 2+k=6;2+k=-6 \\ \Rightarrow k=4;k=-8 \\ \text{Substituting these two values one by one in (1), we get} \\ x+2y-2z+4=0\text{and} \\ x+2y-2z-8=0,\text{which are the equations of the required planes.} \end{gathered}$$