What are the equations of the planes that are parallel to the plane #x+2y-2z=1# and two units away from it?

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:
#vec n=<1,2-2>#

The equation of the plane parallel to the original one passing through #P(x_0,y_0,z_0)# is:

#vec n*"< "x-x_0,y-y_0,z-z_0> =0#
#<1,2,-2>*"<"x-x_0,y-y_0,z-z_0> =0#
#x-x_0+2y-2y_0-2z+2z_0=0#
#x+2y-2z-x_0-2y_0+2z_0=0#

Or

#x+2y-2z+d=0# [1]
where #a=1#, #b=2#, #c=-2# and #d=-x_0-2y_0+2z_0#

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=-1/2#
#-> P_1 (0,0,-1/2)#

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=|ax_1+by_1+cz_1+d|/sqrt(a^2 + b^2 + c^2)#
#2=|1*0+2*0+(-2)*(-1/2)+d|/sqrt(1+4+4)#
#|d+1|=2*3# => #|d+1|=6#

First solution:
#d+1=6# => #d=5#
#-> x+2y-2z+5=0#

Second solution:
#d+1=-6# => #d=-7#
#-> x+2y-2z-7=0#