A tetrahedron has vertices at

 $O(0,0,0),A(1,2,1),B(2,1,3) \mathrm{and} C(-1,1,2)$.

Then the angle between the faces $OAB$ and  $ABC$ will be equal to

Let  $n_1$ and $n_2$ be the vectors normal to the faces $OAB$ and $ABC$. Then,  $n_1=OA\times OB=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ 2& 1& 3\end{array}\right|=5\hat{i}-\hat{j}-3\hat{k}$ and  $n_2=AB\times AC=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ -2& -1& 1\end{array}\right|=\hat{i}-5\hat{j}-3\hat{k}$. If $\theta$ is the angle between the faces $OAB$ and $ABC$, then $\cos \theta =\frac{n_1\cdot n_2}{\left|n_1\right|\left|n_2\right|}$

$\Rightarrow \cos \theta =\frac{5+5+9}{\sqrt{25+1+9\sqrt{1+25+9}}}=\frac{19}{35}\Rightarrow \theta ={\cos }^{-1}\left(\frac{19}{35}\right)$