Double 1st eqn: 8x+8y+8z=8 and subtract from 3rd: y+2z=3, so y=3-2z.
Substitute for y in 2nd and 3rd eqns: 12x+13(3-2z)+14z=15, 12x+39-12z=15, 12(x-z)=-24, x-z=-2, x=z-2.
Divide 1st eqn by 4: x+y+z=1 and substitute for x and y: z-2+3-2z+z=1, 1=1, implying many solutions.
Check: set z=0, then x=-2, y=3. 1st eqn valid; 2nd eqn valid; 3rd eqn valid. Set z=2, then x=0, y=-1. 1st eqn valid; 2nd eqn valid; 3rd eqn valid.
So the system is 3-variable, has many solutions and given z, x=z-2 and y=3-2z.