Every homogeneous system of 3 equations in
3 unknowns has solution (x,y,z) = (0,0,0).
But sometimes homogeneous systems have other
solutions.  So we go through Gauss elimination:
 
 x - 2y + 3z = 0
3x - 7y - 4z = 0
4x - 4y +  z = 0

Make this augmented matrix of coefficients:

[1 -2  3 | 0]
[3 -7 -4 | 0]
[4 -4  1 | 0]

The idea is to get 0's where the three red
numbers are.

[1 -2  3 | 0]
[3 -7 -4 | 0]
[4 -4  1 | 0]

[1 -2  3 | 0]
[3 -7 -4 | 0]
[4 -2  1 | 0]

Get a 0 where the red 3 is by multiplying row 1 by
-3 and adding it to 1 times row 2, but restore row 1:

-3×[1 -2  3 | 0]
 1×[3 -7 -4 | 0]
   [4 -2  1 | 0]

[1  -2   3 | 0]
[0  -1 -13 | 0]
[4  -2   1 | 0] 

Get a 0 where the red 4 is by multiplying row 1 by
-4 and adding it to 1 times row 3, but restore row 1:

-4×[1  -2   3 | 0]
   [0  -1 -13 | 0]
 1×[4  -2   1 | 0]

[1  -2   3 | 0]
[0  -1 -13 | 0]
[0   6 -11 | 0]

Get a 0 where the red 6 is by multiplying row 2 by
6 and adding it to 1 times row 3, but restore row 2:

  [1  -2   3 | 0]
6×[0  -1 -13 | 0]
1×[0   6 -11 | 0] 
  
[1  -2   3 | 0]
[0  -1 -13 | 0]
[0   0 -89 | 0]

Get a 1 where the red -1 is by dividing row 2 thru by -1
Get a 1 where the red -89 is by dividing row 3 thru by -89
 
[1  -2   3 | 0]
[0   1  13 | 0]
[0   0   1 | 0]

Convert back into a system of equations:

1x - 2y +  3z = 0
0x + 1y + 13z = 0
0x + 0y +  1z = 0

Erase the first 0 term of the 2nd equation,
Erase the first two 0 terms of the third equation:
Erase the 1 coefficients

 x + 2y +  3z = 0
      y + 13z = 0
            z = 0

The last equation tells us the only value of z is 0

Substitute in the 2nd equation:

     y + 13(0) = 0
             y = 0

Substitute z = 0 and y = 0 in the first equation:

  x + 2(0) + 3(0) = 0
        x + 0 + 0 = 0
                x = 0

So there is but one solution,
the obvious one.

(x, y, z) = (0, 0, 0)

Edwin