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### Problem Analysis
The given system of equations is:

\[
\begin{cases}
-2x + y + z = a \\
x - 2y + z = b \\
x + y - 2z = c
\end{cases}
\]

We need to show that this system has no solution unless \( a + b + c = 0 \), and if \( a + b + c = 0 \), the system has infinitely many solutions. Additionally, we need to find the solutions when \( a = 1 \), \( b = 1 \), and \( c = -2 \).

### Solution Approach

1. **Combine Equations to Derive Consistency Condition**:
   Let's add all three equations:
   \[
   (-2x + y + z) + (x - 2y + z) + (x + y - 2z) = a + b + c
   \]

   Simplify:
   \[
   -2x + x + x + y - 2y + y + z + z - 2z = a + b + c
   \]

   This reduces to:
   \[
   0 = a + b + c
   \]

   This implies that the system has no solution unless \( a + b + c = 0 \).

2. **Find the Solutions When \( a + b + c = 0 \)**:
   When \( a + b + c = 0 \), the system can be reduced to dependent equations, which means it has infinitely many solutions.

3. **Specific Case: \( a = 1 \), \( b = 1 \), \( c = -2 \)**:
   Substitute these values into the system:

   \[
   \begin{cases}
   -2x + y + z = 1 \\
   x - 2y + z = 1 \\
   x + y - 2z = -2
   \end{cases}
   \]

   Since \( 1 + 1 - 2 = 0 \), the condition \( a + b + c = 0 \) is satisfied.

### Solving for Infinitely Many Solutions

To solve this system, we can express one variable in terms of the others. Let's solve the system using row reduction or substitution.

Let's rewrite the equations as follows:

1. Equation 1: \(-2x + y + z = 1\)
2. Equation 2: \(x - 2y + z = 1\)
3. Equation 3: \(x + y - 2z = -2\)

Using substitution or elimination, we express variables in terms of a free variable, typically \( z \).

From Equation 1:
\[
y = 1 + 2x - z
\]

Substitute this into Equation 2:
\[
x - 2(1 + 2x - z) + z = 1
\]

Simplify:
\[
x - 2 - 4x + 2z + z = 1 \implies -3x + 3z = 3 \implies x = z - 1
\]

Now substitute \( x = z - 1 \) into \( y = 1 + 2x - z \):
\[
y = 1 + 2(z - 1) - z = z - 1
\]

So the solution in terms of \( z \) is:
\[
x = z - 1, \quad y = z - 1, \quad z = z
\]

### Final Solution:
The system has infinitely many solutions given by:
\[
(x, y, z) = (z - 1, z - 1, z)
\]

Let me know if you need further clarification or details!

### Follow-up Questions:
1. How can we use row reduction to confirm the solution set?
2. What would change if \( a + b + c \neq 0 \)?
3. Can the solution be parametrized differently with a different free variable?
4. How do we interpret the infinite solutions geometrically?
5. What happens to the system if any coefficients change?

**Tip**: For systems with infinitely many solutions, expressing variables in terms of a free variable helps to understand the solution space clearly.

Show that the system of equations: -2x + y + z = a, x - 2y + z = b, x + y - 2z = c have no solution unless a + b + c = 0, in which case they have infinitely many solutions. Find these solutions when a = 1, b = 1, c = -2.

Learn how to solve a system of linear equations where the condition a + b + c = 0 determines whether the system has no solution or infinitely many solutions. This problem demonstrates parametric solutions and consistency in linear systems.

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