The system of linear equations `x+lambday-z=0` `lambdax-y-z=0` `x+y-lambdaz=0` has a non-trivial solution for : (1) infinitely many values of `lambda` . (2) exactly one value of `lambda` . (3) exactly two values of `lambda` . (4) exactly three values of `lambda` .

Options

A. Exactly one value of `lambda`
B. Exactly two values of `lambda`
C. Exactly three values of `lambda`
D. Infinitely many values of `lambda`

Explanation

To determine the values of \(\lambda\) for which the given system of linear equations has a non-trivial solution, we can express the equations in matrix form and find the determinant of the coefficient matrix. If the determinant is zero, the system has a non-trivial solution.

### Step-by-step Solution:

1. **Write the system of equations in matrix form**:
The given equations are:
\[ x + \lambda y - z = 0 \quad (1) \]
\[ \lambda x - y - z = 0 \quad (2) \]
\[ x + y - \lambda z = 0 \quad (3) \]

The corresponding coefficient matrix \(A\) is:
\[ A = \begin{bmatrix} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -\lambda \end{bmatrix} \]

2. **Calculate the determinant of the matrix**:
We need to find \(\text{det}(A)\) and set it to zero for a non-trivial solution:
\[ \text{det}(A) = \begin{vmatrix} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -\lambda \end{vmatrix} \]

Expanding the determinant along the first row:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} -1 & -1 \\ 1 & -\lambda \end{vmatrix} - \lambda \cdot \begin{vmatrix} \lambda & -1 \\ 1 & -\lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} \lambda & -1 \\ 1 & 1 \end{vmatrix} \]

Calculate each of these 2x2 determinants:
- For the first determinant:
\[ \begin{vmatrix} -1 & -1 \\ 1 & -\lambda \end{vmatrix} = (-1)(-\lambda) - (-1)(1) = \lambda + 1 \]

- For the second determinant:
\[ \begin{vmatrix} \lambda & -1 \\ 1 & -\lambda \end{vmatrix} = \lambda(-\lambda) - (-1)(1) = -\lambda^2 + 1 = 1 - \lambda^2 \]

- For the third determinant:
\[ \begin{vmatrix} \lambda & -1 \\ 1 & 1 \end{vmatrix} = \lambda(1) - (-1)(1) = \lambda + 1 \]

Substituting back into the determinant:
\[ \text{det}(A) = 1(\lambda + 1) - \lambda(1 - \lambda^2) - 1(\lambda + 1) \]
Simplifying:
\[ = \lambda + 1 - \lambda + \lambda^3 - \lambda - 1 = \lambda^3 - \lambda \]

3. **Set the determinant to zero**:
\[ \lambda^3 - \lambda = 0 \]
Factor out \(\lambda\):
\[ \lambda(\lambda^2 - 1) = 0 \]
Further factor:
\[ \lambda(\lambda - 1)(\lambda + 1) = 0 \]

4. **Find the values of \(\lambda\)**:
The solutions are:
\[ \lambda = 0, \quad \lambda = 1, \quad \lambda = -1 \]

Thus, there are **three values of \(\lambda\)** for which the system has a non-trivial solution.

### Conclusion:
The correct option is (4) exactly three values of \(\lambda\).