Given a regular tetrahedron $S-ABC$ with height $SO=3$, and the side length of the base is 6. A perpendicular line is drawn from point $A$ to the opposite face $SBC$, with the foot of the perpendicular being $O'$. On $AO'$, take a point $P$ such that $\frac{AP}{PO'}=8$. Find the area of the section passing through point $P$ and parallel to the base.
As shown in the figure, since $S-ABC$ is a regular triangular pyramid, $O$ is the centroid of $\triangle ABC$. Connecting $AO$ and extending it to intersect $BC$ at $D$, because $D$ is the midpoint of $BC$, $BC \perp$ plane $SAD$, and $AO' \perp BC$, so $AO'$ lies on plane $SAD$, thus $O'$ must be on $DS$, hence
$$
\begin{array}{l}
AD=6 \times \frac{\sqrt{3}}{2}=3 \sqrt{3}, \\
OD=\frac{1}{3} AD=\sqrt{3} .
\end{array}
$$
$$
SD=\sqrt{3^{2}+(\sqrt{3})^{2}}=\sqrt{12} .
$$

And
$$
\frac{O'D}{AD}=\frac{OD}{SD},
$$

then
$$
O'D=\frac{OD}{SD} \cdot AD=\frac{3}{2} \sqrt{3} .
$$

Let the intersection point of the section through $P$ and parallel to the base with $SD$ be $O''$, then $\frac{O''D}{O'D}=\frac{AP}{O'A}=\frac{8}{9}$, i.e., $O''D=\frac{4}{3} \sqrt{3}$,
$$
\begin{array}{l}
SO''=SD-O''D=\frac{2}{3} \sqrt{3}, \\
\frac{SO''^2}{SD^2}=\frac{1}{9},
\end{array}
$$

thus the area of the required section is
$$
\frac{1}{9} S_{\triangle ABC}=\sqrt{3} \text {. }
$$