17.200 Let $S$ be the set of points on two rays that intersect at an angle of $120^{\circ}$, and let $P$ be a fixed point on the angle bisector and inside the angle. Consider all distinct equilateral triangles $\triangle PQR$, where $Q$ and $R$ are points in $S$. (Points $Q$ and $R$ can be on the same ray; also, in this problem, it is agreed that: “swapping the names of $Q$ and $R$ does not produce a different triangle”) Such triangles
(A) there are exactly 2.
(B) there are exactly 3.
(C) there are exactly 7.
(D) there are exactly 15.
(E) there are more than 15.
[Solution 1] As shown in the figure, let $O$ be the vertex of the $120^{\circ}$ angle, and $A, B, C$ be three points on $S$, satisfying $\angle A P O=60^{\circ}$, thus $\triangle A P O$ is an equilateral triangle.

Point $B$ is any point on $A O$, and point $C$ is on the other side, with $A B=O C$.
Since
$$
\begin{array}{l}
P A=P O, A B=O C, \\
\angle P A B=60^{\circ}=\angle P O C,
\end{array}
$$

we have $\triangle P A B \cong \triangle P O C$.
Therefore, $\angle A P B=\angle O P C$,
which implies $\angle A P O=\angle B P C$.
Since $B P=C P, \angle B P C=60^{\circ}$, $\triangle B P C$ is also an equilateral triangle.
Since $B$ is any point on $A O$, this indicates that there are infinitely many equilateral triangles that satisfy the condition.
Therefore, the answer is $(E)$.
[Solution 2] Draw any circle with $O P$ as the chord, intersecting the sides of the angle at points $R, Q$. Then $\triangle P Q R$ is an equilateral triangle.

Notice that $\angle P R Q=\angle P O Q=60^{\circ}, \angle P Q R=$ $\angle P O R=60^{\circ}$.
Thus, such triangles can be constructed infinitely.
Therefore, the answer is $(E)$.