Answer
The system of equations is inconsistent and has no solutions.
Work Step by Step
We need to solve the given system of equations:
$$2x -2y+3z =6 ~~~(1) \\ 4x -3y+2z =0 ~~~(2) \\ -2x +3y-7z=1 ~~~~(3) $$
Add equations $(1)$ and $(3)$ to eliminate $x$. $$2x -2y+3z-2x+3y-7z=6+1 \\ y-4z=7 ~~~ (4)$$
We can see notice that $x$ is eliminated.
Now, multiply equation $(3)$ by $2$ and then add it to equation $(2)$ to obtain:
$$(4x -3y+2z) +2(-2x +3y -7z) =0+ 2(1) \\4x -3y+2z -4x +6y -14z =0+ 2 \\ 3y +2z -14z =2 \\ 3y-12z=2 \\ y-4z=\dfrac{2}{3} ~~~(5)$$
Now, subtract equation $(5)$ from equation $(4)$. $$y-4z-y+4z = 7-\dfrac{2}{3} \\ 0=\dfrac{19}{3}$$
Since $0$ never equals $\dfrac{19}{3}$, the system of equations is inconsistent and has no solutions.