12. In the tetrahedron $ABCD$, there is a point $O$ inside such that the lines $AO, BO, CO, DO$ intersect the faces $BCD, ACD, ABD, ABC$ at points $A_1, B_1, C_1, D_1$ respectively, and $\frac{AO}{OA_1}=\frac{BO}{OB_1}=\frac{CO}{OC_1}=\frac{DO}{OD_1}=k$. Then $k=$ $\qquad$ .
12. $3 \frac{V_{A B C D}}{V_{O B C D}}=\frac{A A_{1}}{O A_{1}}=\frac{A O}{O A_{1}}+\frac{O A_{1}}{O A_{1}}=1+k$, thus $\frac{V_{O B C D}}{V_{A B C D}}=\frac{1}{1+k}$.

Similarly, $\frac{V_{(X \cdot D A}}{V_{A B C D}}=\frac{V_{O D A B}}{V_{A B C D}}=\frac{V_{O A B C}}{V_{A B C D}}=\frac{1}{1+k}$.
Since $V_{O B C D}+V_{\triangle C D A}+V_{O D A B}+V_{O A B C}=V_{A B C D}$,
we have $\frac{4}{1+k}=1$, hence $k=3$.