If two lines, whose equations are $\frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-1}{\mathrm{\lambda }}\text{ and }\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}-2}{3}$ lie in the same plane. Then, the value of ${\sin }^{-1}\left(\sin \mathrm{\lambda }\right)$ is

These lines lie in the same plane. So, both are coplanar. 

$\therefore \left|\begin{array}{ccc}2& 3& \mathrm{\lambda }\\ 3& 2& 3\\ 1& -1& -1\end{array}\right|=0$

$\begin{array}{l}\Rightarrow 2+18-5\mathrm{\lambda }=0\Rightarrow 5\mathrm{\lambda }=20\Rightarrow \mathrm{\lambda }=4\\ {\sin }^{-1} (\sin \mathrm{\lambda })={\sin }^{-1} (\sin 4)={\sin }^{-1} \sin (\mathrm{\pi }-4)\\ \Rightarrow \mathrm{\lambda }=\mathrm{\pi }-4\end{array}$