x–3y+3z = 10
2x+ y–3z = –3
 x–2y+ z =  6

Add the first two equations above
term by term and the x's will
cancel:

 x–3y+3z = 10
2x+ y–3z = –3
-------------
3x-2y    =  7
 
That has z eliminated. So we 
eliminate z from the 2nd and
3rd equations at the top:

 x–3y+3z = 10
 x–2y+ z =  6

We need to multiply x-2y+z=6
through by -3 to make the
z's cancel when we add them 
term by term:

  x–3y+3z =  10
-3x+6y-3z = -18
---------------
-2x+3y    =  -8

So we have a system of only
two equations in two unknowns:

 3x-2y =  7
-2x+3y = -8

To make the y's cancel multiply
the first one by 3 and the second 
one through by 2

 9x-6y =  21
-4x+6y = -16
------------
 5x    =   5
     x = 1

Substitute x = 1 in

  3x-2y = 7
3(1)-2y = 7   
   3-2y = 7
    -2y = 4
      y = -2

Substitute x = 1 and y = -2 in one
of the original equations to find
z.

     x–3y+3z = 10
(1)-3(-2)+3z = 10
  1 + 6 + 3z = 10
      7 + 3z = 10
          3z = 3
           z = 1

Solution (x,y,z) = (1,-2,1)

Edwin