3. In two regular tetrahedrons $A-O B C$ and $D-O B C$ with coincident bases, $M$ and $N$ are the centroids of $\triangle A D C$ and $\triangle B D C$ respectively. Let $\overrightarrow{O A}=\boldsymbol{a}, \overrightarrow{O B}=\boldsymbol{b}, \overrightarrow{O C}=\boldsymbol{c}$. If point $P$ satisfies $\overrightarrow{O P}=x \boldsymbol{a}+y \boldsymbol{b}+z \boldsymbol{c}, \overrightarrow{M P}=2 \overrightarrow{P N}$, then the real number $9 x+81 y$ $+729 z=$ $\qquad$
Take $O$ as the origin and the line $O B$ as the $x$-axis, establish a spatial rectangular coordinate system as shown in Figure 13.
Let $B(1,0,0)$. Then
$$
\begin{array}{l}
C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right), \\
A\left(\frac{1}{2}, \frac{\sqrt{3}}{6}, \frac{\sqrt{6}}{3}\right), \\
D\left(\frac{1}{2}, \frac{\sqrt{3}}{6},-\frac{\sqrt{6}}{3}\right) .
\end{array}
$$

From the problem and $\overrightarrow{M P}=2 \overrightarrow{P N}$, we know
$$
P\left(\frac{11}{18}, \frac{13 \sqrt{3}}{54},-\frac{2 \sqrt{6}}{27}\right) \text {. }
$$

Thus $\overrightarrow{O P}=x a+y b+z c$
$$
\begin{array}{l}
\Rightarrow x=-\frac{2}{9}, y=\frac{4}{9}, z=\frac{5}{9} \\
\Rightarrow 9 x+81 y+729 z=439 .
\end{array}
$$