A tetrahedron has vertices at $0(0,0, 0), A(1,2,1), B(2, 1, 3)$ and $C(-1, 1, 2).$ Then the angle between the faces $OAB$ and $ABC$ will be:

Angle between two plane faces is equal to the angle between the normals $n_1$ and $n_2$ to the planes. $n_1,$the normal to the face $OAB$ is given by 

$\overrightarrow{OA}\times \overrightarrow{OB}=\left|\begin{array}{l}\mathbf{i}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathbf{j}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathbf{k}\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\end{array}\right|=5\mathbf{i}-\mathbf{j}-3\mathbf{k}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{i}\mathbf{\right)}$

$n_2,$the normal to the face $ABC,$ is given by $\overrightarrow{AB}\times \overrightarrow{AC}.$

${\mathbf{n}}_2=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& -1& 2\\ -2& -1& 1\end{array}\right|=\mathbf{i}-5\mathbf{j}-3\mathbf{k}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{ii}\mathbf{\right)}$

If $\theta$be the angle between $n_1$ and $n_2$

Then $\cos \theta =\frac{{\mathbf{n}}_1\cdot {\mathbf{n}}_2}{\left|{\mathbf{n}}_1\parallel {\mathbf{n}}_2\right|}=\frac{5.1+5+9}{\sqrt{35}\sqrt{35}}$

$\cos \theta =\frac{19}{35}\Rightarrow \theta ={\cos }^{-1}\left(\frac{19}{35}\right).$