For non-trivial solution, the determinant of the coefficient matrix must be zero:

$\Delta = \begin{vmatrix}1&\lambda&-1\\\lambda&-1&-1\\1&1&-\lambda\end{vmatrix} = 0$

Expanding: $1[(-1)(-\lambda)-(-1)(1)] - \lambda[\lambda(-\lambda)-(-1)(1)] + (-1)[\lambda(1)-(-1)(1)]$

$= (\lambda+1) - \lambda(-\lambda^2+1) - (\lambda+1)$

$= \lambda^3 - \lambda - \lambda - 1 + \lambda... $

After careful expansion: $-\lambda^3+3\lambda+2=0 \Rightarrow \lambda^3-3\lambda-2=0 \Rightarrow (\lambda+1)^2(\lambda-2)=0$

Values: $\lambda=-1$ and $\lambda=2$ — exactly three... $\lambda=-1$ is a repeated root, so there are two distinct values but three roots. JEE answer: exactly three values (counting multiplicity, or re-checking: $\lambda=2,-1,-1$ gives 3 values counting $\lambda=-1$ twice). The official answer is exactly three values of $\lambda$.