We are given the system:

x—y+2z=3

2x+z=1

3x+2y+z=4

First, we eliminate y. Multiply (1) by 2 to obtain (4):

2x-2y+4z=6

Add (3) and (4) to obtain (5): 5x+5z= 10

or by dividing by 5, x+z=2

Next, we eliminate 2. Subtract (2) and (5) then solve for x:

x=-1

Solve for 2 using (5):

-l+z=2

z=3

Solve for y using (1):

—1-y+2(3)=3

-y+5=3

-y=-2

y=2

So, the solution of the system is:

(-1,2,3)

x—y+2z=3

2x+z=1

3x+2y+z=4

First, we eliminate y. Multiply (1) by 2 to obtain (4):

2x-2y+4z=6

Add (3) and (4) to obtain (5): 5x+5z= 10

or by dividing by 5, x+z=2

Next, we eliminate 2. Subtract (2) and (5) then solve for x:

x=-1

Solve for 2 using (5):

-l+z=2

z=3

Solve for y using (1):

—1-y+2(3)=3

-y+5=3

-y=-2

y=2

So, the solution of the system is:

(-1,2,3)