Question

Find the equations of the planes parallel to the plane x − 2y + 2z − 3 = 0 and which are at a unit distance from the point (1, 1, 1).

Answer 1

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