12. A polyhedron, except for one vertex, the sum of the angles at the vertices of each face of the other vertices is $5160^{\circ}$, then the sum of the angles at the vertices of all faces of the polyhedron is $\qquad$ .
12. $5400^{\circ}$

Let the number of $k$-sided faces in a polyhedron be $F_{k}$, then the total sum of all face angles of the polyhedron is
$$
S=180^{\circ}\left(F_{3}+2 F_{4}+3 F_{5}+\cdots\right) \text {, and } 2 E=3 F_{3}+4 F_{4}+5 F_{5}+\cdots \text {, }
$$

where $E$ is the sum of the edges of the polyhedron, so $\frac{S}{180^{\circ}}+2 E=4 F_{3}+6 F_{4}+8 F_{5}+\cdots$
Therefore, $S=360^{\circ} m, m \in \mathbf{N}^{*}$, and let $S^{\prime}$ represent the sum of the uncounted face angles, $0<S^{\prime}<360^{\circ}$, and
$$
360^{\circ} m-S^{\prime}=5160^{\circ}=14 \times 360^{\circ}+120^{\circ}
$$

Thus, $m=5, S^{\prime}=240^{\circ}$, so $S=5160^{\circ}+240^{\circ}=5400^{\circ}$.