3. In two regular tetrahedrons $A-OBC$ and $D-OBC$ with their bases coinciding, $M$ and $N$ are the centroids of $\triangle ADC$ and $\triangle BDC$ respectively. Let $\overrightarrow{OA}=\boldsymbol{a}, \overrightarrow{OB}=\boldsymbol{b}, \overrightarrow{OC}=\boldsymbol{c}$. If point $P$ satisfies $\overrightarrow{OP}=x \boldsymbol{a}+y \boldsymbol{b}+z \boldsymbol{c}, \overrightarrow{MP}=2 \overrightarrow{PN}$, then the real number $9 x+81 y+729 z=$ $\qquad$
(Adapted from the 2016 National High School Mathematics League Zhejiang Regional Preliminary Contest)
Take $O$ as the origin and the line $O B$ as the $x$-axis, establishing 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 $P\left(\frac{11}{18}, \frac{13 \sqrt{3}}{54},-\frac{2 \sqrt{6}}{27}\right)$.
Thus $\overrightarrow{O P}=x a+y b+z c$
$\Rightarrow x=-\frac{2}{9}, y=\frac{4}{9}, z=\frac{5}{9}$
$$
\Rightarrow 9 x+81 y+729 z=439 \text {. }
$$