12. Given two regular tetrahedrons $A O B C$ and square $D O B C$, $M$ and $N$ are the centroids of $\triangle A D C$ and $\triangle B D C$ respectively. Let $\overrightarrow{O A}=a, \overrightarrow{O B}=\boldsymbol{b}, \overrightarrow{O C}=c$. If point $P$ satisfies
$$
\overrightarrow{O P}=x a+y b+z c, \overrightarrow{M P}=2 \overrightarrow{P N} \text {, }
$$

then the real numbers $x=$ $\qquad$ ,$y=$ $\qquad$ ,$z=$ $\qquad$
12. $x=-\frac{2}{9}, y=\frac{4}{9}, z=\frac{5}{9}$.

Let point $A$ project onto plane $OBC$ at point $H$.
$$
\begin{array}{l}
\text { Then } \overrightarrow{O H}=\frac{2}{3} \times \frac{1}{2}(\overrightarrow{O B}+\overrightarrow{O C})=\frac{1}{3}(b+c) \\
\Rightarrow \overrightarrow{A H}=\overrightarrow{O H}-\overrightarrow{O A}=\frac{1}{3}(b+c-3 a) \\
\Rightarrow \overrightarrow{A D}=2 \overrightarrow{A H}=\frac{2}{3}(b+c-3 a) . \\
\text { Also, } \overrightarrow{A M}=\frac{2}{3} \times \frac{1}{2}(\overrightarrow{A D}+\overrightarrow{A C}) \\
=\frac{1}{9}(-9 a+2 b+5 c), \\
\text { Therefore, } \overrightarrow{O M}=\overrightarrow{O A}+\overrightarrow{A M}=\frac{1}{9}(2 b+5 c) .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
\overrightarrow{B N}=\frac{2}{3} \times \frac{1}{2}(\overrightarrow{B C}+\overrightarrow{B D}) \\
=\frac{1}{3}(\overrightarrow{O C}-\overrightarrow{O B}+\overrightarrow{A D}-\overrightarrow{A B}) \\
=\frac{1}{9}(-3 a+4 b+5 c), \\
\overrightarrow{O N}=\overrightarrow{O B}+\overrightarrow{B N}=\frac{1}{9}(-3 a+5 b+5 c) \\
\text { Given } \overrightarrow{M P}=2 \overrightarrow{P N} \\
\Rightarrow 3 \overrightarrow{O P}=\overrightarrow{O M}+2 \overrightarrow{O N} \\
\Rightarrow x a+y b+z c=\frac{1}{9}(-2 a+4 b+5 c) \\
\Rightarrow x=-\frac{2}{9}, y=\frac{4}{9}, z=\frac{5}{9} .
\end{array}
$$