Example 6 Given a regular tetrahedron $S-A B C$ with height $S O=3$, and the side length of the base is 6. Draw a perpendicular from point $A$ to the opposite face $S B C$, with the foot of the perpendicular being $O^{\prime}$. Take a point $P$ on $A O^{\prime}$ such that $A P: P O^{\prime}=8$. Find the area of the section passing through point $P$ and parallel to the base.
Solve As shown in Figure 17-10, since $S-A B C$ is a regular triangular pyramid, $O$ is the center of $\triangle A B C$. Connect $A O$ and extend it to intersect $B C$ at point $D$.

Since $D$ is the midpoint of $B C$, $B C \perp$ plane $S A D$.
Thus, plane $S A D \perp$ plane $S B C$, so $O^{\prime}$ must lie on $D S$.
$$
A D=3 \sqrt{3}, O D=\frac{1}{3} A D=\sqrt{3}, S D=\sqrt{3^{2}+(\sqrt{3})^{2}}=2 \sqrt{3}
$$

By $O^{\prime} D: A D=O D: S D$, we have $O^{\prime} D=\frac{O D}{S D} \cdot A D=\frac{3}{2} \sqrt{3}$
Let the intersection point of the section passing through point $P$ and parallel to the base with $S D$ be ()$^{\prime \prime}$.
Then $P()^{\prime \prime} / / A D$.
Thus, $\quad \frac{O^{\prime} D}{O^{\prime} D}=\frac{P A}{O^{\prime} A}=\frac{8}{9}, O^{\prime \prime} D=\frac{8}{9} O^{\prime} D=\frac{4}{3} \sqrt{3}$
$$
S()^{\prime \prime}=S D-O^{\prime \prime} D=\frac{2}{3} \sqrt{3}
$$

Therefore, $\mathrm{SO}^{\prime \prime 2}: \mathrm{SD}^{2}=1: 9$
Thus, the area of the required section is $\frac{1}{9} S_{\triangle A B C}=\frac{1}{9} \times \frac{\sqrt{3}}{4} \times 6^{2}=\sqrt{3}$