There are a number of ways of solving this system. Here’s one.
Label the equations:
(1) 4y+3z=-7
(2) 12x-3y+z=-7
(3) -6x+2y-2z=12
(4): (2)+2(3): y-3z=17
(5): (4)+(1): 5y=10, so y=2
From (1) 3z=-7-4y=-7-8=-15, so z=-15/3=-5.
From (2) 12x=-7+3y-z=-7+6+5=4, x=4/12=⅓.
Solution is x=⅓, y=2, z=-5.