* 2. On the three sides of $\triangle A B C$, take points $P_{1}, P_{2}$, $P_{3}, P_{4}, P_{5}, P_{6}, \cdots$, such that $P_{1}, P_{4}, P_{7}, \cdots$ are on $A C$, $P_{2}, P_{5}, P_{8}, \cdots$ are on $A B$, and $P_{3}, P_{6}, P_{9}, \cdots$ are on $B C$, and $A P_{1}=A P_{2}$, $B P_{2}=B P_{3}, C P_{3}=C P_{4}, A P_{4}=A P_{5}, B P_{5}=$ $B P_{6}, \cdots$. Then $\left|P_{2} P_{1994}\right|=$ $\qquad$
2. 0

Construct the incircle of $\triangle ABC$. By the property of tangent segments, we can prove that $P_{n+6}=P_{6}$. Therefore, $P_{1994}=P_{2}$, which gives $P_{2} P_{1994}=$ 0. Now, let's handle this using complex numbers.

Set up the complex plane, and let each point's letter represent the complex number at that point. We have
$$
\begin{array}{l}
P_{2}-A=e^{A i}\left(P_{1}-A\right), \\
P_{5}-A=e^{A i}\left(P_{4}-A\right) .
\end{array}
$$

Subtracting, $P_{5}-P_{2}=e^{A i}\left(P_{4}-P_{1}\right)$.
Similarly, $P_{6}-P_{3}=e^{R i}\left(P_{5}-P_{2}\right)$
$$
=e^{(A-B) i}\left(P_{4}-P_{1}\right),
$$
$$
\begin{aligned}
P_{7}-P_{4} & =e^{C_{i}}\left(P_{6}-P_{3}\right) \\
& =e^{(A+B+C) i}\left(P_{4}-P_{1}\right) \\
& =e^{\pi i}\left(P_{4}-P_{1}\right)=-\left(P_{4}-P_{1}\right),
\end{aligned}
$$

we get $P_{7}=P_{1}$:
Similarly, $P_{n+6}=P_{n}$.
Since $1994=6 \times 332+2$,
we know $P_{1994}=P_{2}$.
Thus, $\left|P_{2} P_{1994}\right|=0$.