Write the equation of the set of points at distance d=4 from…

Write the equation of the set of points at distance $d=4$ from the plane $2x+2y+z-8=0$.

Explanation of Method

The set of points at distance 4 from a given plane consists of two parallel planes.

Step 1: Solution

\[ \frac{|2x+2y+z-8|}{3}=4\ \Longrightarrow\ 2x+2y+z-8=\pm12 . \]

\[2x+2y+z-20=0\quad\text{and}\quad 2x+2y+z+4=0\]

These are two planes parallel to the given one, one on each side. Check: $\tfrac{|-8+20|}{3}=\tfrac{|-8-4|}{3}=4$. The two planes are $8$ apart. The union can also be written as $(2x+2y+z-8)^2=144$.

$4$ $4$

$2x+2y+z-20=0$
$2x+2y+z-8=0$
$2x+2y+z+4=0$

The locus: two planes parallel to the given one (gray), at distance $4$ on either side.

Result

\[\boxed{\,2x+2y+z-20=0\quad\text{and}\quad 2x+2y+z+4=0\,}\]