$17 \cdot 161$ As shown in the figure, on the sides $BC, CA$, and $AB$ of the equilateral $\triangle ABC$, there are internal division points $D, E, F$, dividing the sides in the ratio $2:(n-2)$ (where $n>4$). The segments $AD, BE, CF$ intersect to form $\triangle PQR$, whose area is $\frac{1}{7}$ of the area of $\triangle ABC$. What is the value of $n$?
(A) 5.
(B) 6.
(C) 7.
(D) 8.
[Solution] As shown in the figure, $B D: D C=C E: E A=A F: F B=2:(n-2)$.
Notice that line $B P E$ intersects the sides or their extensions of $\triangle A D C$ at $P, B, E$. By Menelaus' theorem, we have
$$
\frac{A P}{P D} \cdot \frac{D B}{B C} \cdot \frac{C E}{E A}=\frac{A P}{P D} \cdot \frac{2}{n} \cdot \frac{2}{n-2}=1 .
$$

Therefore, $\quad \frac{A P}{P D}=\frac{n(n-2)}{4}, \frac{A P}{A D}=\frac{n(n-2)}{n(n-2)+4}$.
Thus, $S_{\triangle A B P}=\frac{n(n-2)}{n(n-2)+4} S_{\triangle A B D}=\frac{n(n-2)}{n(n-2)+4} \cdot \frac{2}{n} S_{\triangle A B C}$
$$
=\frac{2(n-2)}{n(n-2)+4} S_{\triangle A B C} \text {. }
$$

Similarly,
$$
S_{\triangle B C Q}=S_{\triangle C A R}=\frac{2(n-2)}{n(n-2)+4} S_{\triangle A B C} .
$$

Therefore, $S_{\triangle A B P}+S_{\triangle B C Q}+S_{\triangle C A R}=\frac{6(n-2)}{n(n-2)+4} S_{\triangle A B C}$.
Given that $\frac{6(n-2)}{n(n-2)+4}=\frac{6}{7}$, solving for $n$ yields $n=6$. Hence, the answer is $(B)$.