Part A.
The sum of the first two equations is (2x +3y -z) +(x -3y +2z) = (5) +(-6), or 3x +z = -1.
Part B.
A good next-step in the process is to eliminate y from another equation, such as by adding the second equation to 3 times the third one. Doing that would give two equations in the two unknowns x and z, making a simpler system to solve.
... (x -3y +2z) +3(3x +y -4z) = (-6) +3(-8)
... 10x -10z = -30
Dividing this equation by 10 gives
... x - z = -3
Part C.
Adding the equation just found to the one of Part A gives ...
... (x -z) + (3x +z) = (-3) +(-1)
... 4x = -4
... x = -1
... z = x + 3 = 2
... y = -8 +4z -3x . . . . from the third original equation
... y = -8 +8 -3(-1) = 3
The solution is (x, y, z) = (-1, 3, 2).
Yes, it works when you test it.
... 2(-1) +3(3) -(2) = 5
... (-1) -3(3) +2(2) = -6
... 3(-1) +(3) -4(2) = -8