We are given the following system of equations:

$1. \quad x + 2y + 3z = 0$ $2. \quad 2x + y + 3z = 0$ $3. \quad 3x + 2z = 1$

We need to solve this system for $z$.

Step 1: Solve for $y$ in terms of $x$ and $z$

From equation (1): $x + 2y + 3z = 0$ Rearrange to express $y$ in terms of $x$ and $z$: $2y = -x - 3z$ $y = \frac{-x - 3z}{2}$

Step 2: Substitute $y$ into the second equation

Substitute $y = \frac{-x - 3z}{2}$ into equation (2): $2x + y + 3z = 0$ $2x + \frac{-x - 3z}{2} + 3z = 0$

Multiply through by 2 to eliminate the fraction: $4x - x - 3z + 6z = 0$ Simplify: $3x + 3z = 0$ $x + z = 0$ $x = -z$

Step 3: Substitute $x = -z$ into the third equation

Substitute $x = -z$ into equation (3): $3x + 2z = 1$ $3(-z) + 2z = 1$ Simplify: $-3z + 2z = 1$ $-z = 1$ $z = -1$

Final Answer:

The value of $z$ is $\boxed{-1}$.

Would you like further clarification or details? Here are some related questions for deeper understanding:

1. What is the value of $x$ in this system?
2. How do we solve for $y$ using the value of $z$?
3. How can we verify the solution by substituting values back into the original equations?
4. Can this system be solved using matrices or another method?
5. What does the solution to this system tell us about the nature of the equations?

Tip: When solving systems of equations, it's often helpful to express one variable in terms of others to simplify the process.