17・157 As shown in the figure, the area of the equilateral $\triangle ABC$ is $1$. $B P=\frac{1}{3} B C$, $A Q=B Q$, and $P R \perp A C$ at $R$. Then the area of $\triangle P R Q$ is
(A) $\frac{7}{12}$.
(B) $\frac{1}{2}$.
(C) $\frac{1}{3}$.
(D) $\frac{3}{8}$.
(E) $\frac{5}{18}$.
[Solution] It is easy to prove that $R C=\frac{1}{2} P C=\frac{1}{3} A C$ (by drawing $B D \perp A C$ at $D$, and using the properties of similar triangles). Therefore, we have
$$
\begin{aligned}
& S_{\triangle Q B P}=\frac{1}{6}, \quad S_{\triangle P R C}=\frac{2}{9}, \quad S_{\triangle A Q R}=\frac{1}{3} . \\
\therefore \quad & S_{\triangle P Q R}=1-\frac{1}{6}-\frac{2}{9}-\frac{1}{3}=\frac{5}{18} .
\end{aligned}
$$

Therefore, the answer is $(E)$.