For some real number $\lambda$, if the area of the triangle having $\overline{a}=3\overline{i}-\overline{j}+\lambda \overline{k}$ and  $\overline{b}=\lambda \overline{i}+\overline{j}-3\overline{k}$ as two of its sides is $\frac{\sqrt{195}}{2}$, then the number of distinct possible values of $\lambda$ is

Area of triangle $=\frac{1}{2}\left|\overline{a}\times \overline{b}\right|$

$\frac{\sqrt{195}}{2}=\frac{1}{2}\left|\begin{array}{ccc}\overline{i}& \overline{j}& \overline{k}\\ 3& -1& \lambda \\ \lambda & 1& -3\end{array}\right|$

$\frac{1}{2}\sqrt{{\left(3-\lambda \right)}^2+{\left(9+{\lambda }^2\right)}^2+{\left(3+\lambda \right)}^2}=\frac{\sqrt{195}}{2}$

Square on both sides
 (After Simplification)
${\lambda }^4+20{\lambda }^2+99=195$

${\lambda }^4+20{\lambda }^2+99-195=0$

${\lambda }^2\left({\lambda }^2+20\right)=96$

$\lambda =2\left(or\right)\lambda =-2$

There are two distinct possible values of  $"\lambda "$