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

For co-planar vectors, we have
\(\begin{aligned}
& \left|\begin{array}{ccc}
-\lambda^2 & 1 & 1 \\
1 & -\lambda^2 & 1 \\
1 & 1 & -\lambda^2
\end{array}\right|=0 \\
& \therefore \quad-\lambda^2\left(\lambda^4-1\right)-1\left(-\lambda^2-1\right)+1\left(1+\lambda^2\right)=0 \\
& \therefore \quad-\lambda^6+\lambda^2+\lambda^2+1+1+\lambda^2=0 \\
& \therefore \quad \lambda^6-3 \lambda^2-2=0 \\
& \therefore \quad t^3-3 t-2=0 \\
& \ldots\left[\text { For } \mathrm{t}=\lambda^2\right] \\
& \therefore \quad(\mathrm{t}+1)\left(\mathrm{t}^2-\mathrm{t}-2\right)=0 \\
& \therefore \quad(\mathrm{t}+1)(\mathrm{t}+1)(\mathrm{t}-2)=0 \\
& \therefore \quad t=-1 \text { or } t=2 \\
& \text { i.e., } \lambda^2=-1 \text { or } \lambda^2=2
\end{aligned}\)
for real values of \(\lambda\), we get \(\lambda= \pm \sqrt{2}\), two values.