To solve the system of linear equations:

\[
\begin{align*}
1.\ & x + 3y - z = 2 \\
2.\ & x + y - z = 0 \\
3.\ & 3x + 2y - 3z = -1
\end{align*}
\]

We can use the method of elimination or substitution. Here, I'll use the elimination method.

### Step 1: Subtract Equation 2 from Equation 1 to eliminate \(x\):

\[
(x + 3y - z) - (x + y - z) = 2 - 0
\]

This simplifies to:

\[
2y = 2 \quad \Rightarrow \quad y = 1
\]

### Step 2: Substitute \(y = 1\) into Equation 2:

\[
x + 1 - z = 0 \quad \Rightarrow \quad x - z = -1 \quad \text{(Equation 4)}
\]

### Step 3: Substitute \(y = 1\) into Equation 3:

\[
3x + 2(1) - 3z = -1
\]

This simplifies to:

\[
3x + 2 - 3z = -1 \quad \Rightarrow \quad 3x - 3z = -3 \quad \Rightarrow \quad x - z = -1 \quad \text{(Equation 5)}
\]

### Step 4: Compare Equation 4 and Equation 5:

Both Equation 4 and Equation 5 are identical:

\[
x - z = -1
\]

This implies that \(x = z - 1\).

### Step 5: Substitute \(x = z - 1\) and \(y = 1\) back into any of the original equations to solve for \(z\). Using Equation 2:

\[
(z - 1) + 1 - z = 0
\]

This simplifies to:

\[
0 = 0
\]

This means that \(z\) can take any value, and the system is dependent with infinitely many solutions.

### General Solution:
Given that \(z = t\), where \(t\) is any real number, the solution set is:

\[
\begin{align*}
x & = t - 1 \\
y & = 1 \\
z & = t
\end{align*}
\]

### Final Answer:
The solution is \(x = t - 1\), \(y = 1\), and \(z = t\), where \(t\) is any real number.

Would you like more details or have any questions?

Here are some related questions:
1. How do we determine if a system of equations has a unique solution, no solution, or infinitely many solutions?
2. What is the geometric interpretation of a system of linear equations with infinitely many solutions?
3. How can we use matrices to solve a system of linear equations?
4. What is the substitution method for solving systems of equations, and when is it most useful?
5. How does the elimination method compare to the substitution method in terms of efficiency?

**Tip:** When solving systems of equations, always check for consistency by substituting back into the original equations.

Learn how to solve a system of linear equations with infinite solutions using the elimination method. This problem explores the concept of dependent systems in algebra, suitable for high school students.

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