Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists. Show your work.

4x-y+3z = 12
x+4y+6z =-32
5x+3y+9z = 20

To solve the system of equations using Gaussian elimination, we need to eliminate variables by performing row operations on the augmented matrix until it's in triangular form or reduced row-echelon form. Let's start by writing the augmented matrix for the system of equations:

```
[4 -1 3 | 12]
[1 4 6 | -32]
[5 3 9 | 20]
```

Step 1: Convert the matrix to triangular form

To eliminate the x variable from the second and third equations, multiply the first equation by -1 and add it to the second equation. Also, multiply the first equation by -5 and add it to the third equation. The following operations will be performed:

- Row 2 = Row 2 + (-1) * Row 1
- Row 3 = Row 3 + (-5) * Row 1

Resulting matrix:
```
[ 4 -1 3 | 12]
[ 0 5 3 | -44]
[ 0 8 -6 | -58]
```

Step 2: Continue row operations to achieve row-echelon form

To eliminate the y variable from the third equation, multiply the second equation by -8 and add it to the third equation. The following operation will be performed:

- Row 3 = Row 3 + (-8) * Row 2

Resulting matrix:
```
[ 4 -1 3 | 12]
[ 0 5 3 | -44]
[ 0 0 -30 | 290]
```

Step 3: Solve for z

From the final row, we can determine that `-30z = 290`. Solving for z, we get `z = -290/30 = -29/3`.

Step 4: Solve for y

Substituting the value of z in the second row, we have `5y + 3*(-29/3) = -44`. Simplifying, we get `5y - 29 = -44`. Solving for y, we have `5y = -44 + 29`, which results in `y = -15/5 = -3`.

Step 5: Solve for x

Substituting the values of y and z into the first row, we have `4x - (-3) + 3*(-29/3) = 12`. Simplifying, we get `4x + 3 + 29 = 12`. Solving for x, we have `4x = 12 - 3 - 29`, which gives `x = -20/4 = -5`.

Therefore, the solution to the system of equations is x = -5, y = -3, z = -29/3.