As we know that if four points $P, A, B, C$ are coplanar, the vectors $\overrightarrow{PA},\overrightarrow{PB},\overrightarrow{PC}$ must be coplanar.

$\Rightarrow \overrightarrow{PA}.\left(\overrightarrow{PB}\times \overrightarrow{PC}\right)=0$

$\Rightarrow \left|\begin{array}{ccc}-{\lambda }_{ }^2+1& 2& 0\\ 2& -{\lambda }^2+1& 0\\ 2& 2& -{\lambda }^2-1\end{array}\right|=0$

$\Rightarrow \left(-{\lambda }^2+1\right)\left[\left(-{\lambda }^2+1\right)\left(-{\lambda }^2-1\right)-0\right]-2\left[2\left(-{\lambda }^2-1\right)-0\right]+0\left[4-2\left(-{\lambda }^2+1\right)\right]=0$

$\Rightarrow \left(-{\lambda }^2+1\right)\left({\lambda }^4-1\right)+4\left({\lambda }^2+1\right)=0$

$\Rightarrow \left(-{\lambda }^2+1\right)\left({\lambda }^2+1\right)\left({\lambda }^2-1\right)+4\left({\lambda }^2+1\right)=0$

$\Rightarrow \left({\lambda }^2+1\right)\left\{\left(-{\lambda }^2+1\right)\left({\lambda }^2-1\right)+4\right\}=0$

$\Rightarrow -\left({\lambda }^2+1\right)\left({\lambda }^4-2{\lambda }^2+3\right)=0$

$\Rightarrow -\left({\lambda }^2+1\right)\left({\lambda }^2-3\right)\left({\lambda }^2+1\right)=0$

$\Rightarrow \lambda =\pm \sqrt{3}$