Question 785389
When my solution does not check, or looks too complicated, I throw away my prior attempt, take a few deep breaths, and start again from scratch writing as big and neatly as possible on a large, clean piece of paper. I did that with your problem, twice, and the third time was a charm. (If I reviewed what I had done looking for the mistake, I would be most likely to miss it).
 
In {{{system(3x-2y+3z=11,5x+2y-2z=4,-x+y+z=-7)}}},
-x+y+z=-7  looks like the simplest of the three equations.
I am going to use it to eliminate {{{z}}} from the other two.
(The sooner I eliminate {{{z}}} the less likely I am to confuse it with a {{{2}}}).
 
I multiply -x+y+z=-7 times {{{(-3)}}} and add it to 3x-2y+3z=11:
3x -2y  +3z  =  11
3x -3y  -3z  =  21
______________
6x -5y       =  32
 
I multiply -x+y+z=-7 times {{{2}}} and add it to 5x+2y-2z=4:
 5x +2y  -2z  =  4
-2x +2y  +2z  = -14
______________
3x +4y       =  -10
 
Now I have the equivalent system
{{{system(-x+y+z=-7,6x-5y=32,3x+4y=-10)}}}
I am going to combine 6x-5y=32 and 3x+4y=-10 to get an equation without {{{x}}}.
 
I multiply 3x+4y=-10 times {{{(-2)}}} and add it to 6x-5y=32:
 6x -5y  =  32
-6x -8y  =  20
___________
-13y  =  52  ---------> y = 52 / (-13) -----> {{{highlight(y=-4)}}}
 
Now I have the equivalent system
{{{system(-x+y+z=-7,3x+4y=-10, y=-4)}}}
 
Substituting y=-4 into 3x+4y=-10, I get
3x+4(-4)=-10  ------> 3x-16 = -10 ------> 3x=-10+16 ----> 3x=6 ---> x=6/3 ---> {{{highlight(x=2)}}}
 
Now I have the equivalent system
{{{system(-x+y+z=-7,x=2, y=-4)}}}
 
Substituting x=2 and y= -4 into -x+y+z=-7, I get
-2 + (-4) +z = -7 ---> -2-4 +z = -7 ---> -6 + z = -7 --> z=-7+6 ---> {{{highlight(z=1)}}}