17・160 As shown in the figure, $C D, A E$ and $B F$ are each $\frac{1}{3}$ of the side lengths, accordingly, $A N_{2}: N_{2} N_{1}: N_{1} D=3: 3: 1$. Similarly, $B E$ and $C F$ are the same, then the area of triangle $N_{1} N_{2} N_{3}$ is
(A) $\frac{1}{10} \triangle A B C$.
(B) $\frac{1}{9} \triangle A B C$.
(C) $\frac{1}{7} \triangle A B C$.
(D) $\frac{1}{6} \triangle A B C$.
(E) None of the above.
(3rd American High School Mathematics Examination, 1952)
[Solution] Given $C D=\frac{1}{3} B C$, then $S_{\triangle A C D}=\frac{1}{3} S_{\triangle A B C}$.
Also, $A N_{2}: N_{2} N_{1}: N_{1} D=3: 3: 1$, so $N_{1} D=\frac{1}{7} A D$.
Thus, $S_{\triangle C D N_{1}}=\frac{1}{7} S_{\triangle A C D}=\frac{1}{7} \cdot \frac{1}{3} S_{\triangle A B C}=\frac{1}{21} S_{\triangle A B C}$.
Similarly, $S_{\triangle A E V_{2}}=S_{\triangle B F V_{3}}=\frac{1}{21} S_{\triangle A B C}$.
Therefore, $S_{\triangle N_{1} V_{2} V_{3}}=S_{\triangle A B C}-\left[S_{\triangle A C D}+S_{\triangle A B E}+S_{\triangle C B F}\right]+S_{\triangle C D N_{1}}$
$$
\begin{aligned}
& +S_{\triangle A E V_{2}}+S_{\triangle B E V_{3}} \\
= & S_{\triangle A B C}-3 \cdot \frac{1}{3} S_{\triangle A B C}+3 \cdot \frac{1}{21} S_{\triangle A B C} \\
= & \frac{1}{7} S_{\triangle A B C} .
\end{aligned}
$$

Hence, the answer is $(C)$.