The number of distinct real values of $\lambda$, for which the vectors $-\lambda^{2}\hat{i}+\hat{j}+\hat{k}$, $\hat{i}-\lambda^{2}\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}-\lambda^{2}\hat{k}$ are coplanar, is

$\vec a = -\lambda^2\hat{i}+\hat{j}+\hat{k},\qquad \vec b = \hat{i}-\lambda^2\hat{j}+\hat{k}$ $\vec c = \hat{i}+\hat{j}-\lambda^2\hat{k}$ $\text{To find: number of real } \lambda \text{ with } \vec a,\vec b,\vec c \text{ coplanar}$ $\text{Coplanar} \Leftrightarrow [\vec a\ \vec b\ \vec c] = 0$ $[\vec a\ \vec b\ \vec c] = \begin{vmatrix} -\lambda^2 & 1 & 1\\ 1 & -\lambda^2 & 1\\ 1 & 1 & -\lambda^2 \end{vmatrix} = 0$ $\text{Let } t=-\lambda^2:\ \text{diagonal } t,\ \text{off-diagonal } 1$ $\det = (t-1)^2(t+2)$ $(-\lambda^2-1)^2(2-\lambda^2) = 0$ $\lambda^2+1 = 0 \Rightarrow \text{no real } \lambda$ $2-\lambda^2 = 0 \Rightarrow \lambda = \pm\sqrt{2}$ $\lambda=\sqrt2:\ \vec a=(-2,1,1),\ \vec b=(1,-2,1),\ \vec c=(1,1,-2)$ $\vec a+\vec b+\vec c = \vec 0 \Rightarrow \text{coplanar}$ $\text{Number of real } \lambda = 2 \Rightarrow \text{option }(3)$ $\boxed{2\quad \text{option (3)}}$