$17 \cdot 158$ An equilateral triangle $ABC$ has an area of 36, and points $P, Q, R$ are on $BC$, $AB$, $CA$ respectively, such that $BP=\frac{1}{3}BC$, $AQ=QB$, and $PR \perp AC$. Then the area of $\triangle PQR$ is 
(A) 10.
(B) 12.
(C) 14.
(D) 15.
(China Junior High School Mathematics Correspondence Competition, 1988)
[Solution] Let the side length of equilateral $\triangle A B C$ be $6a$,
then we have $A Q=Q B=3a, B P=2a, P C=4a, C R=2a, R A=4a$.
$$
\begin{array}{ll}
\because & S_{\triangle A B C}=\frac{1}{2}(6a)^{2} \sin 60^{\circ}=36, \\
\therefore & \frac{1}{2} a^{2} \sin 60^{\circ}=1 .
\end{array}
$$

Thus,
$$
\begin{aligned}
& S_{\triangle P Q R}=S_{\triangle A B C}-\left(S_{\triangle A R Q}+S_{\triangle B Q P}+S_{\triangle C P R}\right) \\
= & 36-\left(\frac{1}{2} \cdot 4a \cdot 3a \sin 60^{\circ}+\frac{1}{2} \cdot 3a \cdot 2a \sin 60^{\circ}+\frac{1}{2} \cdot 4a \cdot 2a \sin 60^{\circ}\right) \\
= & 36-26 \cdot \frac{1}{2} a^{2} \sin 60^{\circ}=36-26=10 .
\end{aligned}
$$

Therefore, the answer is $(A)$.