Question

Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?

Answer 1

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
→n=<1,2−2>

The equation of the plane parallel to the original one passing through P(x0,y0,z0)is:

→n⋅< x−x0,y−y0,z−z0>=0
<1,2,−2>⋅<x−x0,y−y0,z−z0>=0
x−x0+2y−2y0−2z+2z0=0
x+2y−2z−x0−2y0+2z0=0

Or

x+2y−2z+d=0 [1]
where a=1, b=2, c=−2 and d=−x0−2y0+2z0

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when x=0 and y=0:
x+2y−2z=1 => 0+2⋅0−2z=1 => z=−12
→P1(0,0,−12)

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping D=2, and d as d itself, we get:

D=|ax1+by1+cz1+d|√a2+b2+c2
2=∣∣1⋅0+2⋅0+(−2)⋅(−12)+d∣∣√1+4+4
|d+1|=2⋅3 => |d+1|=6First solution:
d+1=6 => d=5
→x+2y−2z+5=0Second solution:
d+1=−6 => d=−7
→x+2y−2z−7=0