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**Hint:**If we are given the system of equations like here we are given:

$x + \lambda y - z = 0$

$\lambda x - y - z = 0$

$x + y - \lambda z = 0$

If all variable values are not equal to $0$, then it is a non-trivial solution.

$\left| {\begin{array}{*{20}{c}}

1&\lambda &{ - 1} \\

\lambda &{ - 1}&{ - 1} \\

1&1&{ - \lambda }

\end{array}} \right| = 0$

Get the determinant of coefficients and find the value of $\lambda $.

**Complete step-by-step answer:**

So here we are given the system of equations like here we are given:

$x + \lambda y - z = 0$

$\lambda x - y - z = 0$

$x + y - \lambda z = 0$

And it is a non-trivial solution.

It means that at least one of the variables must not be equal to $0$.

If all are equal to zero, then it is a trivial solution.

So here the determinant of the coefficients as shown must be equal to $0$

$\left| {\begin{array}{*{20}{c}}

1&\lambda &{ - 1} \\

\lambda &{ - 1}&{ - 1} \\

1&1&{ - \lambda }

\end{array}} \right| = 0$

Now solving it, we get

$1(\lambda + 1) - \lambda ( - {\lambda ^2} + 1) - 1(\lambda + 1) = 0$

$(\lambda + 1) - \lambda (1 - {\lambda ^2}) - 1(\lambda + 1) = 0$

We know identity $(a-b) (a+b) = a^2-b^2$

Applying the identity to 2nd term

$(\lambda + 1) + \lambda ({\lambda ^2}-1) - 1(\lambda + 1) = 0$

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

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

Taking $(\lambda + 1)$ and simplifying further we get,

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

So we get

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

So there are three values for which it is a non-trivial solution.

**So, the correct answer is “Option D”.**

**Note:**If set of the linear equations are given to be trivial then variables all are equal to zero but if it is a non-trivial solution then at least one must not be equal to zero.

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