2x - 2y - 4z = -2
 3x - 3y - 6z = -3
-2x + 3y +  z  = 7

It is true that if you try to eliminate any variable in 1&2
everything cancels out.  This means there are infinitely 
many solutions, but we can still find the general solution.
So try to eliminate a letter using 2&3.  Notice that the y's
will cancel if you add 2&3 as they are:

 3x - 3y - 6z = -3
-2x + 3y +  z  = 7
------------------
  x      - 5z  = 4

So solve for  x:

             x = 4 + 5z

Substitute 4 + 5z for x in either 1 or 2. I'll pick 1

     2x - 2y - 4z = -2
2(4+5z) - 2y - 4z = -2 
8 + 10z - 2y - 4z = -2
      8 + 6z - 2y = -2

Solve for y:

              -2y = -10 - 6z
                y =  - 
                y = 5 + 3z

So the general solution is

x = 4 + 5z, y = 5 + 3z,  z = z

often written as a 3D point:

(x, y, z) = (4+5z, 5+3z, z)

There are infinitely many solutions.

To get some of them, substitute arbitrary values for z:

Substitute, say, 1 for z:

(x, y, z) = (4+5z, 5+3z, z) = (4+5·1, 5+3·1, z) = (9, 8, 1)

Substitute, say, -2 for z:

(x, y, z) = (4+5z, 5+3z, z) = (4+5·-2, 5+3·-2, z) = (4-10, 5-6, -2) = (-6, -1, -2)

You can get as many different solutions as you like just by 
substituting arbitrary values for z.

Edwin