Example 5 Given a regular tetrahedron $S-ABC$ with height $SO=3$, and the side length of the base is 6. A perpendicular 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 parallel to the base and passing through point $P$.
(1989, National High School Mathematics Competition)
Solution: As shown in Figure 5, since $S-A B C$ is a regular tetrahedron,
point $O$ is the centroid of $\triangle A B C$.
Connecting $A O$ and extending it to intersect $B C$ at $D$,
since point $D$ is the midpoint of $B C$, and $B C \perp$ plane $S A D$, and $A O^{\prime} \perp B C$, $A O^{\prime}$ lies on plane $S A D$. Therefore, point $O^{\prime}$ must be on $D S$. Thus,
$$
\begin{array}{l}
A D=6 \times \frac{\sqrt{3}}{2}=3 \sqrt{3}, O D=\frac{1}{3} A D=\sqrt{3}, \\
S D=\sqrt{3^{2}+(\sqrt{3})^{2}}=\sqrt{12}=2 \sqrt{3} . \\
\text { Since } \frac{O^{\prime} D}{A D}=\frac{O D}{S D} \text {, then } \\
O^{\prime} D=\frac{O D}{S D} \cdot A D=\frac{3}{2} \sqrt{3} .
\end{array}
$$

Let the intersection point of the plane through point $P$ and parallel to the base with $S D$ be $O^{\prime \prime}$, then $\frac{O^{\prime \prime} D}{O^{\prime} D}=\frac{A P}{O^{\prime} A}=\frac{8}{9}$, i.e., $O^{\prime \prime} D=$ $\frac{4}{3} \sqrt{3}$
$$
S O^{\prime \prime}=S D-O^{\prime \prime} D=\frac{2}{3} \sqrt{3}, \frac{S O^{\prime 2}}{S D^{2}}=\frac{1}{9} \text {. }
$$

Therefore, the area of the required section is
$$
\frac{1}{9} S_{\triangle A B C}=\sqrt{3} .
$$