The system of equations `-2x+y+z=a` `x-2y+z=b` `x+y-2z=c` has

Options

A. no solution if `a+b+c ne 0`
B. unique solution if `a+b+c =0`
C. infinite number of solutions if `a+b+c=0`
D. None of these

Explanation

To solve the system of equations given by:

1. \(-2x + y + z = a\)
2. \(x - 2y + z = b\)
3. \(x + y - 2z = c\)

we will analyze the conditions under which this system has solutions.

### Step 1: Write the system of equations in matrix form
We can express the system of equations in the form of a matrix equation \(AX = B\), where:

\[ A = \begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \]

### Step 2: Calculate the determinant of matrix \(A\)
To determine the nature of the solutions, we need to compute the determinant of matrix \(A\):

\[ \text{det}(A) = \begin{vmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{vmatrix} \]

Calculating the determinant using the formula for a \(3 \times 3\) matrix:

\[ \text{det}(A) = -2 \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} + 1 \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} \]

Calculating the \(2 \times 2\) determinants:

\[ \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} = (-2)(-2) - (1)(1) = 4 - 1 = 3 \]

\[ \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = (1)(-2) - (1)(1) = -2 - 1 = -3 \]

\[ \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-2)(1) = 1 + 2 = 3 \]

Substituting back into the determinant calculation:

\[ \text{det}(A) = -2(3) - 1(-3) + 1(3) = -6 + 3 + 3 = 0 \]

### Step 3: Analyze the determinant
Since \(\text{det}(A) = 0\), this indicates that the system of equations may have either no solutions or infinitely many solutions.

### Step 4: Check the consistency of the equations
To determine the specific case, we can add all three equations:

\[ (-2x + y + z) + (x - 2y + z) + (x + y - 2z) = a + b + c \]

Simplifying the left-hand side:

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

This simplifies to:

\[ 0 = a + b + c \]

### Conclusion
- If \(a + b + c = 0\), then the system has infinitely many solutions.
- If \(a + b + c \neq 0\), then the system has no solutions.

### Final Answer
The system of equations has:
- Infinitely many solutions if \(a + b + c = 0\).
- No solutions if \(a + b + c \neq 0\).