First line: $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{\lambda }$

$A\equiv (3,2,1) \text{is a point on the line}$

Direction ratios are, $\overrightarrow{a}= 2\hat{i}+3\hat{j}+\lambda \hat{k}$

Second line: $\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-2}{3}$

$B\equiv (2,3,2) \text{is a point on the line}$

Direction ratios are, $\overrightarrow{b} = 3\hat{i}+2\hat{j}+3\hat{k}$

$\overrightarrow{AB}= -\hat{i}+\hat{j}+\hat{k} ; \overrightarrow{a}= 2\hat{i}+3\hat{j}+\lambda \hat{k} ; \overrightarrow{b} = 3\hat{i}+2\hat{j}+3\hat{k}$

Since the lines are coplanar, we get

$\left|\begin{array}{ccc}-1& 1& 1\\ 2& 3& \lambda \\ 3& 2& 3\end{array}\right|=0 \Rightarrow \lambda = 4$

Now, $\sin \left(4\right)=\sin \left(\pi -4\right) \text{and }\cos \left(4\right)=\cos \left(2\pi -4\right)$

${\sin }^{-1}\left(\sin 4\right)+{\cos }^{-1}\left(\cos 4\right)={\sin }^{-1}(\sin (\pi -4\left)\right)+{\cos }^{-1}(\cos (2\pi -4\left)\right)$

$=\pi -4+2\pi -4 = 3\pi -8$