Example 10 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 point $P$ such that $AP: PO'=8$. Find the area of the section through point $P$ and parallel to the base.
(1989 National High School League Question)
As shown in Figure 23-17, take the midpoint $D$ of $BC$. By the properties of a regular tetrahedron, we know that $O'$ lies on $SD$. $O$ is the centroid of $\triangle ABC$, so $O$ lies on $AD$. We can examine $\triangle SAD$. Let the plane $\alpha$ through $P$ parallel to the base intersect $SD$ at $R$. Since $\frac{\text{Area of section } \alpha}{S_{\triangle ABC}} = \frac{SR^2}{SD^2}$, we only need to find the ratio of $SR$ to $SD$ to determine the area of the section $\alpha$.

The ratio $SR: SD$ can be obtained by calculating $SD$ and $RD$ specifically. In fact, $SD = \sqrt{SO^2 + OD^2} = 2\sqrt{3}$, $\frac{RD}{O'D} = \frac{PA}{O'A} = \frac{8}{9}$, $RD = \frac{8}{9} O'D$.
Figure 23-17
$$
\text{Also, } \frac{O'D}{AD} = \frac{OD}{SD}, O'D = \frac{OD}{SD} \cdot AD = \frac{3}{2} \sqrt{3}, \text{ thus } RD = \frac{4}{3} \sqrt{3}, SR = SD - RD = \frac{2}{3} \sqrt{3}, \text{ hence the }
$$
required area of the section is $\sqrt{3}$.