In a triangle ABC, If $a^2-b^2-c^2=bc({\lambda }^2-2\lambda -1)$ then 

$\frac{b^2+c^2-a^2}{2ab}=\frac{1}{2}({\lambda }^2-2\lambda -1)$
$$\begin{gathered} \begin{array}{l}\therefore -1\le \frac{1}{2}({\lambda }^2-2\lambda -1)\le 1\\ \Rightarrow -2\le {\lambda }^2-2\lambda -12\\ \Rightarrow -2\le {(\lambda -1)}^2\lambda -2\le 2\\ \Rightarrow \cos A=\frac{-1}{2}({\lambda }^2-2\lambda -1)\\ -1\le \cos A\le 1 \\ \begin{array}{l}\Rightarrow 0\le {(\lambda -1)}^2\le 4\\ \Rightarrow -1\le \lambda \le 3\end{array}\end{array} \end{gathered}$$