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 :

Solution :
We know that three vector are coplanar if their scalar triple product is zero.
\[\begin{array}{l} \Rightarrow \left| \begin{array}{l} - {\lambda ^2}\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\, - {\lambda ^2}\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\,\,\,1\,\,\, - {\lambda ^2} \end{array} \right| = 0\\ {R_1} \to {R_1} + {R_2} + {R_3}\\ \Rightarrow \left| \begin{array}{l} 2 - {\lambda ^2}\,\,\,\,\,2 - {\lambda ^2}\,\,\,\,\,2 - {\lambda ^2}\\ \,\,\,\,\,1\,\,\,\,\,\,\,\,\,\, - {\lambda ^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ \,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {\lambda ^2} \end{array} \right| = 0\\ \Rightarrow \left( {2 - {\lambda ^2}} \right)\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\, - {\lambda ^2}\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\,1\,\,\, - {\lambda ^2} \end{array} \right| = 0\\ \Rightarrow \left( {2 - {\lambda ^2}} \right)\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 0\,\,\,\, - \left( {1 + {\lambda ^2}} \right)\,\,\,\,\,0\\ 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\, - \left( {1 + {\lambda ^2}} \right) \end{array} \right| = 0\\ \left( {{R_2} - {R_1},\,\,{R_3} - {R_1}} \right)\\ \Rightarrow \left( {2 - {\lambda ^2}} \right){\left( {1 + {\lambda ^2}} \right)^2} = 0\\ \Rightarrow \lambda = \pm \sqrt 2 \end{array}\]
$$\therefore $$ Two real solutions.