Answer
No solution
Work Step by Step
The first equation: $x+3y-z=12$
Solve for x: $\rightarrow x=-3y+z+12$
Substitute for z in the second equation:
$2x+4y-2z=6$
$2(-3y+z+12)+4y-2z=6$
$-9y+3z+36+4y-2z=6$
$-2y+z=-30$ (1)
Substitute for z in the third equation:
$-x - 2y + z = -6$
$-(-3y+z+12)- 2y + z = -6$
$y=6$
It is impossible for $y$ to have two different values, so the system is inconsistent and has no solution.
