This would be staggeringly tedious to show on this platform. But, I can give you the starting matrix, and the row operations to resolve it. I have to confess there is a shorter series of row operations that get you to the answer, but I elected this method to avoid fractional values. (Just because of the tedium and cramped writing involved. This has 10 row operations, the optimal method uses 9)
⌈3 -4 -6 | -19⌉
|-3 2 -4 | -21|
⌊ 5 -1 2 | 0⌋
R1 + R2 → R2
R3 - R1 → R3
R1 - R3 → R1
R3 - 2R1 → R3
(-1/2)R2 → R2
R3 - 17R2 → R3
(-1/49)R3 → R3
R2 - 5R3 → R2
R1 + 14R3 → R1
R1 + 7R2 → R1
which will give a final matrix of
⌈1 0 0 | -3⌉
|0 1 0 | -5|
⌊0 0 1 | 5⌋
so x = -3, y = -5, and z = 5