The third equation suggests you might want to use the first two to eliminate y. That way, you could end up with two equations in x and z, making it easy to solve for those variables. To do that (eliminate y), you add twice the second equation to the first:
... (x +2y -4z) +2(2x -y +2z) = (4) +2(3)
... 5x = 10 . . . . . . . . it also works out that z is eliminated, a bonus.
... x = 2 . . . . . . . . . . divide by 5
Now, you can substitute this into the 3rd equation to find z.
.. 3·2 -8z = -2
... -8z = -8 . . . . . . . subtract 6
... z = 1 . . . . . . . . . . divide by -8
Using the second equation, we can find y.
... 2(2) -y +2(1) = 3
...6 - y = 3 . . . . . . . . . simplify
... y = 3 . . . . . . . . . . . add y-3
(x, y, z) = (2, 3, 1)