What is the distance between the planes -6x+9y-9z=27 and 2x-3y+3z=12 ?

To calculate a distance between two planes, the planes
must be parallel. We'll verify if the given planes are
parallel.

We'll divide the first equation -6x+9y-9z=27 by
3:

-2x + 3y - 3z = 9

We'll
multiply by -!:

2x - 3y + 3z =
-9

We notice that the coefficients of the first equation of
the plane are the same with the coefficients of the second plane, therefore the planes
are parallel.

The distance between the given planes is the
distance from a point located in a plane to the other
plane.

Now, we'll choose a point located in the first
plane: M(0,0,2).

The distance formula
is:

d(M,P2) = |2*0 - 3*0 + 3*2 + 9|/sqrt(2^2 + (-3)^2 +
3^2)

d(M,P2) =
9/sqrt22

d(M,P2) =
9sqrt22/22

The requested distance between the
given planes is d(M,P2) = 9sqrt22/22.