(1)        x +  y +  z =  -5
(2)       2x + 3y - 2z =   8
(3)        x -  y + 4z = -21

We can eliminate y by adding 
  (1) and (3) term by term

(1)        x +  y +  z =  -5
(3)        x -  y + 4z = -21
----------------------------
(4)       2x      + 5z = -26

We can eliminate y from (2) and (3)
by multiplying (3) by -3  and adding
to (2)

         3x - 3y + 12z = -63 
(2)      2x + 3y -  2z =   8
----------------------------
         5x      + 10z = -55

And we can divide that through by 5:

(5)       x + 2z = -11

So now we have the 2 by 2 system 
of (4) and (5)

(4)      2x + 5z = -26
(5)       x + 2z = -11

Multiply (5) by -2 and add to (4)

        -2x - 4z =  22
(4)      2x + 5z = -26
         -------------
(6)            z =  -4

Substitute z = -4 in (5)

(5)       x + 2z = -11
       x + 2(-4) = -11
           x - 8 = -11
               x = -3

Substitute x = -3 and z = -4 in (1)

(1)         x + y + z = -5
        -3 + y + (-4) = -5 
                y - 7 = -5
                    y =  2

So solution is (x,y,z) = (-3,2,-4)

Edwin