Question 360033
Remark : (inspired by a comment)
 
Note that the RHS is not linear but affine, hence different from eigenvalue-like problems (1x, -1y, 2z)*, hence the last column cannot be put on the in diagonal on the LHS : 
 
 
 
2 -1 +1 1   | -2last
1 -1 +1 -1  | -last
1 -2 +1 2
 
0 3 -1 -3  | -3mid
0 1 0 -3
1 -2 1 2
 
0 0 -1 6
0 1 0 -3
1 -2 1 2  | +2mid + first
 
0 0 -1 6
0 1 0 -3
1 0 0 2
 
=> x = 2, y = -3, z = -6
 
Verification : 
 
2x-y+z = 4 + 3 - 6 = 7 - 6 = 1
x-y+z = 2 + 3 -6 = 5 - 6 = -1 
x-2y+z = 2 + 6 -6 = 2