7. (1989 National High School League Question) 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 $\frac{AP}{PQ}=8$. Find the area of the section parallel to the base passing through point $P$.
7. Since $S-A B C$ is a regular tetrahedron, $O$ is the centroid of $\triangle A B C$. Connect $A O$ and extend it to intersect $B C$ at $D$. Since $D$ is the midpoint of $B C$, $B C \perp$ plane $S A D$, and $A O^{\prime} \perp B C$, $A O^{\prime}$ lies on plane $S A D$, so $O^{\prime}$ must be on $D S$. Thus, $A D=6 \cdot \frac{\sqrt{3}}{2}=$ $3 \sqrt{3}, O D=\frac{1}{3} A D=\sqrt{3}, S D=\sqrt{3^{2}+(\sqrt{3})^{2}}=\sqrt{12}$. Also, $\frac{O^{\prime} D}{A D}=\frac{O D}{S D}$, so $O^{\prime} D=\frac{O D}{S D} \cdot A D=\frac{\sqrt{3}}{\sqrt{12}} \cdot 3 \sqrt{3}=$ $\frac{3}{2} \sqrt{3}$. Let the intersection point of the plane through $P$ parallel to the base with $S D$ be $O^{\prime \prime}$, then $\frac{O^{\prime} D}{O D}=\frac{A P}{O^{\prime} A}=\frac{8}{9}$.
Thus, $O^{\prime} D=\frac{8}{9} O^{\prime} D=\frac{8}{9} \cdot \frac{3}{2} \sqrt{3}=\frac{4}{3} \sqrt{3}$, and $\frac{S O^{\prime 2}}{S D^{2}}=\frac{\left(\frac{2}{3} \sqrt{3}\right)^{2}}{(\sqrt{12})^{2}}=\frac{1}{9}$. Therefore, the area of the required section is $\frac{1}{9} S_{\triangle M x}=$ $\frac{1}{9} \cdot \frac{1}{2} \cdot 6 \cdot 3 \sqrt{3}=\sqrt{3}$