One way to solve this is to use substitution.
From ① x=13-4z and we can put this into the other two equations: ② 4(13-4z)-2y+z=7, 52-16z-2y+z=7, -15z-2y=-45 or 15z+2y=45.
And ③ 2(13-4z)-2y-7z=-19, 26-8z-2y-7z=-19, -15z-2y=-45 or 15z+2y=45. This is the same equation as ②!
Therefore we cannot find unique solutions for z and y, and that implies no unique solution for x either.
To prove this point let z=0, then x=13 and y=45/2. Sure enough, this satisfies all the original equations.
Now let z=1, then x=9 and ② becomes 36-2y+1=7, y=15. ③ becomes 18-2y-7=-19. And y=15 again!
Let z=2, then x=5 and y=15/2.
We can go on doing this and we will have as many solutions as we wish! So the equations can only show a relationship, not a unique solution. This relationship can also be represented graphically.