3. Given $B D: D C=C E: E A=A F: F B=3:(n-3)$, applying Menelaus' theorem to $\triangle A D C$ and the transversal $B P E$, we have $\frac{A P}{P D} \cdot \frac{D B}{B C} \cdot \frac{C E}{E A}=\frac{A P}{P D} \cdot \frac{3}{n} \cdot \frac{3}{n-3}=1$, thus $\frac{A P}{P D}=\frac{n(n-3)}{9}, \frac{A P}{A D}=\frac{n(n-3)}{n(n-3)+9}$, hence $S_{\triangle A B P}=\frac{n(n-3)}{n(n-3)+9}$. $S_{\triangle A B D}=\frac{3(n-3)}{n(n-3)+9} \cdot S_{\triangle A B C}$. Similarly, we can find $S_{\triangle B C Q}, S_{\triangle C A R}$, thus $S_{\triangle A B P}+S_{\triangle B C Q}+S_{\triangle C A R}=\frac{9(n-3)}{n(n-3)+9}$. $S_{\triangle A B C}$. From the problem, $\frac{9(n-3)}{n(n-3)+9}=\frac{45}{49}$, which simplifies to $5 n^{2}-64 n+192=0$, noting $n>6$, solving yields $n=8$.
3. On the sides $BC$, $CA$, $AB$ of the equilateral $\triangle ABC$, there are internal division points $D$, $E$, $F$ dividing the sides in the ratio $3:(n-3)$ (where $n>6$). The segments $AD$, $BE$, $CF$ intersect to form $\triangle PQR$ (where $BE$ intersects $AD$ at $P$ and $FC$ at $Q$). When the area of $\triangle PQR$ is $\frac{4}{49}$ of the area of $\triangle ABC$, find the value of $n$.
(1992 Japan Mathematical Olympiad Preliminary)