The system of linear equations
$$x + \lambda y - z = 0$$
$$\lambda x - y - z = 0$$
$$x + y - \lambda z = 0$$
has a non-trivial solution for

We are given the homogeneous linear system

$$x+\lambda y-z=0$$

$$\lambda x-y-z=0$$

$$x+y-\lambda z=0$$

A homogeneous system has a non-trivial (i.e. not all variables zero) solution precisely when the determinant of its coefficient matrix is zero. We first write the coefficient matrix:

$$A=\begin{pmatrix} 1 & \lambda & -1\\[4pt] \lambda & -1 & -1\\[4pt] 1 & 1 & -\lambda \end{pmatrix}$$

The condition for a non-trivial solution is $$\det A = 0.$$ Now we expand this determinant along the first row. The formula for expansion is

$$\det A = a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13},$$

where each $$C_{1j}=(-1)^{1+j}M_{1j}$$ and $$M_{1j}$$ is the minor obtained by deleting row 1 and column j.

We compute each part step by step.

First term (column 1): $$a_{11}=1.$$ We remove the first row and first column to get the minor matrix $$\begin{pmatrix} -1 & -1\\ 1 & -\lambda \end{pmatrix},$$ whose determinant is $$(-1)(-\lambda)-(-1)(1)=\lambda+1.$$
So the contribution is $$1\cdot(\lambda+1)=\lambda+1.$$

Second term (column 2): $$a_{12}=\lambda,$$ and the sign factor is $$(-1)^{1+2}=-1.$$ The minor matrix after deleting row 1 and column 2 is $$\begin{pmatrix} \lambda & -1\\ 1 & -\lambda \end{pmatrix},$$ whose determinant is $$\lambda(-\lambda)-(-1)(1)= -\lambda^{2}+1=1-\lambda^{2}.$$
Hence the contribution is $$-\lambda\,(1-\lambda^{2})=-\lambda+\lambda^{3}.$$

Third term (column 3): $$a_{13}=-1,$$ and the sign factor is $$(-1)^{1+3}=+1.$$ Deleting row 1 and column 3 leaves $$\begin{pmatrix} \lambda & -1\\ 1 & 1 \end{pmatrix},$$ whose determinant equals $$\lambda(1)-(-1)(1)=\lambda+1.$$
Thus the contribution is $$(-1)\cdot(\lambda+1)=-\lambda-1.$$

Adding the three contributions we get

$$\det A=(\lambda+1)+\bigl(-\lambda+\lambda^{3}\bigr)+\bigl(-\lambda-1\bigr).$$

Now we combine like terms carefully:

$$\det A=\lambda+1-\lambda+\lambda^{3}-\lambda-1 =\lambda^{3}-\lambda.$$

Factorising,

$$\det A=\lambda(\lambda^{2}-1)=\lambda(\lambda-1)(\lambda+1).$$

For a non-trivial solution we set this determinant to zero:

$$\lambda(\lambda-1)(\lambda+1)=0.$$

This equation is satisfied when

$$\lambda=0,\quad \lambda=1,\quad \lambda=-1.$$

Thus there are exactly three distinct values of $$\lambda$$ (namely $$-1,\,0,\,1$$) for which the system admits a non-trivial solution.

Hence, the correct answer is Option B.