If l + m + n = 0, then the system of equations
`-2x + y + z = l`
`x - 2y + z = m`
`x + y - 2z = n`
has

Options

A. a trivial solution
B. no solution
C. a unique solution
D. infinitely many solutions

Explanation

To solve the given system of equations, we will follow these steps:

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \(-2x + y + z = l\)
2. \(x - 2y + z = m\)
3. \(x + y - 2z = n\)

We can express this system in matrix form as:
\[ A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} l \\ m \\ n \end{pmatrix} \]
where \(A\) is the coefficient matrix:
\[ A = \begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{pmatrix} \]
and the vector of variables is:
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} \]
and the constant matrix is:
\[ \begin{pmatrix} l \\ m \\ n \end{pmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix \(A\)
To determine the nature of the solutions, we need to calculate the determinant of matrix \(A\):
\[ \text{det}(A) = -2 \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} + 1 \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} = (-2)(-2) - (1)(1) = 4 - 1 = 3 \]
\[ \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = (1)(-2) - (1)(1) = -2 - 1 = -3 \]
\[ \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-2)(1) = 1 + 2 = 3 \]
Now substituting back:
\[ \text{det}(A) = -2(3) - 1(-3) + 1(3) = -6 + 3 + 3 = 0 \]

### Step 3: Analyze the determinant
Since \(\text{det}(A) = 0\), the system of equations is either inconsistent or has infinitely many solutions.

### Step 4: Check for infinite solutions
To determine whether there are infinite solutions, we need to check the condition of the equations. We will find the adjoint of matrix \(A\) and check if \( \text{adj}(A) \cdot \begin{pmatrix} l \\ m \\ n \end{pmatrix} = 0 \).

### Step 5: Calculate the adjoint of matrix \(A\)
The adjoint of a matrix is the transpose of the cofactor matrix. We compute the cofactors:
- For \(a_{11}\): \(\text{Cofactor} = \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} = 3\)
- For \(a_{12}\): \(\text{Cofactor} = -\begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = 3\)
- For \(a_{13}\): \(\text{Cofactor} = \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} = 3\)
- For \(a_{21}\): \(\text{Cofactor} = -\begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = -3\)
- For \(a_{22}\): \(\text{Cofactor} = \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} = 3\)
- For \(a_{23}\): \(\text{Cofactor} = -\begin{vmatrix} -2 & 1 \\ 1 & 1 \end{vmatrix} = 1\)
- For \(a_{31}\): \(\text{Cofactor} = \begin{vmatrix} 1 & -2 \\ -2 & 1 \end{vmatrix} = 3\)
- For \(a_{32}\): \(\text{Cofactor} = -\begin{vmatrix} -2 & 1 \\ 1 & 1 \end{vmatrix} = -1\)
- For \(a_{33}\): \(\text{Cofactor} = \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} = 3\)

Thus, the adjoint matrix is:
\[ \text{adj}(A) = \begin{pmatrix} 3 & 3 & 3 \\ -3 & 3 & -1 \\ 3 & -1 & 3 \end{pmatrix} \]

### Step 6: Multiply adjoint \(A\) with the constant matrix
Now we compute:
\[ \text{adj}(A) \cdot \begin{pmatrix} l \\ m \\ n \end{pmatrix} = \begin{pmatrix} 3 & 3 & 3 \\ -3 & 3 & -1 \\ 3 & -1 & 3 \end{pmatrix} \begin{pmatrix} l \\ m \\ n \end{pmatrix} \]
This results in:
\[ \begin{pmatrix} 3l + 3m + 3n \\ -3l + 3m - n \\ 3l - m + 3n \end{pmatrix} \]
Given that \(l + m + n = 0\), we can substitute:
\[ 3(l + m + n) = 3 \cdot 0 = 0 \]
Thus, the first component is 0.

### Conclusion
Since \(\text{adj}(A) \cdot \begin{pmatrix} l \\ m \\ n \end{pmatrix} = 0\), we conclude that the system has infinitely many solutions.

### Final Answer
The system of equations has infinitely many solutions.