It looks like you're talking about row-reducing an augmented matrix to solve the system of equations. Your answer is almost correct. The last row should read 0, 0, 1, 2/7.
The given system translates to
Eliminate x from the last two rows by combining -2 (row 2) and row 1, and -2 (row 3) and row 1; that is,
(2x - 3y + z) - 2 (x - y + 2z) = 2 - 2 (2)
2x - 3y + z - 2x + 2y - 4z = 2 - 4
-y - 3z = -2
and
(2x - 3y + z) - 2 (x + 2y - 3z) = 2 - 2 (4)
2x - 3y + z - 2x - 4y + 6z = 2 - 8
-7y + 7z = -6
In augmented matrix form, this step yields
I'll omit these details in the remaining steps.
Eliminate y from the last row by combining -7 (row 2) and row 3 :
Multiply the last row by 1/28 :
Eliminate z from the second row by combining 3 (row 3) and row 2 :
Multiply the second row by -1 :
Eliminate y and z from the first row by combining 3 (row 2), -1 (row 3), and row 1 :
Multiply the first row by 1/2 :