A tetrahedron has vertices at O(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

$cos^{-1}(\frac{19}{35})$

Let $\vec{n_1}$ and $\vec{n_2}$ be the  vectors normal to the faces OAB and ABC. Then,

$\vec{n_1}= \vec{OA}× \vec{OB}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & 2 & 1\\2 & 1 & 3\end{vmatrix}= 5\hat{i} - \hat{j} - 3\hat{k}$

and,

$\vec{n_2}= \vec{AB}× \vec{AC}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & -1 & 2\\-2 & -1 & 1\end{vmatrix}= \hat{i} -5 \hat{j} - 3\hat{k}$

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

$cos θ = \frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}$

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