To solve this system of equations, we can use the method of substitution or elimination.

Using elimination method:

Multiply the second equation by 2 to make the coefficient of x the same in the first and second equations:

2x + 4y - 6z = 28

Add the first and modified second equation:

2x + 3y + 2z = 9
2x + 4y - 6z = 28

7y - 4z = 37

Now, we have two equations:

7y - 4z = 37
3x + 4y + z = 16

Multiply the first equation by 3 and add to the third equation:

21y - 12z = 111
3x + 4y + z = 16

3x + 25y = 127

Rearrange this equation to solve for x:

3x = 127 - 25y
x = (127 - 25y) / 3

Substitute this value of x back into one of the original equations, like the first one:

(127 - 25y) / 3 + 2y = 9
(127 - 25y) + 6y = 27
127 - 19y = 27
-19y = -100
y = 5.26

Now that we have the value of y, substitute back into one of the equations to find the values of x and z:

x = (127 - 25*5.26) / 3 = 2.69
Substitute y = 5.26 and solve for x:

2.69 + 3*5.26 + 2z = 9
2.69 + 15.78 + 2z = 9
18.47 + 2z = 9
2z = -9.47
z = -4.74

Therefore, the solution to the system of equations is:
x = 2.69, y = 5.26, z = -4.74