**System of Equations:**

The given system of equations is:
x + 2y + 3z = 2 ...(1)
2x + 3y + z = -1 ...(2)
x - y + z = -2 ...(3)

**Matrix Form:**

The system of equations can be represented in matrix form as AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

The coefficient matrix A is:
A = [[1, 2, 3], [2, 3, 1], [1, -1, 1]]

The variable matrix X is:
X = [[x], [y], [z]]

The constant matrix B is:
B = [[2], [-1], [-2]]

**Augmented Matrix:**

The augmented matrix [A|B] is formed by combining A and B as follows:
[A|B] = [[1, 2, 3, 2], [2, 3, 1, -1], [1, -1, 1, -2]]

**Row Operations:**

Performing row operations on the augmented matrix [A|B] to transform it into row-echelon form:

1. R2 = R2 - 2R1:
[[1, 2, 3, 2], [0, -1, -5, -5], [1, -1, 1, -2]]

2. R3 = R3 - R1:
[[1, 2, 3, 2], [0, -1, -5, -5], [0, -3, -2, -4]]

3. R3 = 3R3 + R2:
[[1, 2, 3, 2], [0, -1, -5, -5], [0, 0, -17, -19]]

**Row-Echelon Form:**

The row-echelon form of the augmented matrix [A|B] is:
[[1, 2, 3, 2], [0, -1, -5, -5], [0, 0, -17, -19]]

**Back Substitution:**

Using back substitution, we can solve for the variables.

From the third row of the row-echelon form, we have:
-17z = -19
Solving this equation, we get z = 1

Substituting z = 1 in the second row of the row-echelon form, we have:
-y - 5(1) = -5
Simplifying this equation, we get y = 0

Substituting y = 0 and z = 1 in the first row of the row-echelon form, we have:
x + 2(0) + 3(1) = 2
Simplifying this equation, we get x = -1

**Solution:**

The solution to the given system of equations is:
x = -1, y = 0, z = 1