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

$\text{ Let }{\overrightarrow{n}}_1$ and ${\overrightarrow{n}}_2$ be the vectors normal to the faces OAB and ABC. Then

${\overrightarrow{n}}_1=\overrightarrow{OA}\times \overrightarrow{OB}=\left|\begin{array}{lll}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ 2& 1& 3\end{array}\right|=5\hat{i}-\hat{j}-3\hat{k}$

and ${\overrightarrow{n}}_2=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{rrr}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ -2& -1& 1\end{array}\right|=\hat{i}-5\hat{j}-3\hat{k}$

If 0 is the angle between the faces OAB and ABC, then 

$\cos \theta =\frac{{\overrightarrow{n}}_1·{\overrightarrow{n}}_2}{\left|{\overrightarrow{n}}_1\right|\left|{\overrightarrow{n}}_2\right|}\Rightarrow \cos \theta =\frac{5+5+9}{\sqrt{25+1+9}\sqrt{1+25+9}}=\frac{19}{35}$

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