3. Let $P_{1}$ be a point on the side $A B$ of the equilateral $\triangle A B C$, draw a perpendicular from $P_{1}$ to the side $B C$, with the foot of the perpendicular being $Q_{1}$, draw a perpendicular from $Q_{1}$ to the side $C A$, with the foot of the perpendicular being $R_{1}$, draw a perpendicular from $R_{1}$ to the side $A B$, with the foot of the perpendicular being $P_{2}$, and so on, obtaining points $Q_{2}, R_{2}, P_{3}, Q_{3}, R_{3}$, $\cdots$, as $n \rightarrow \infty$, which point does $P_{n}$ approach infinitely?
3. As shown in the figure, let $B P_{n}=a_{n}, A B=a$,

then $B P_{n+1}=a_{n+1}$.
$$
\begin{array}{l}
\text { By } B P_{n}=a_{n} \Rightarrow B Q_{n}=\frac{1}{2} a_{n} \Rightarrow Q_{n} C=a-\frac{1}{2} a_{n} \\
\Rightarrow C R_{n}=\frac{1}{2}\left(a-\frac{1}{2} a_{n}\right) \Rightarrow A R_{n} \\
=a-C R_{n}=\frac{1}{2} a+\frac{1}{4} a_{n} \\
\Rightarrow A P_{n+1}=\frac{1}{2} A R_{n}=\frac{1}{4} a+\frac{1}{8} a_{n} \\
\Rightarrow B P_{n+1}=a_{n+1}=a-A P_{n+1} \\
=a-\left(\frac{1}{4} a+\frac{1}{8} a_{n}\right)=\frac{3}{4} a-\frac{1}{8} a_{n}
\end{array}
$$

Thus, we obtain the recurrence relation $a_{n+1}=\frac{3}{4} a-\frac{1}{8} a_{n}(n \geqslant 1)$, and it is easy to get $a_{n}=\frac{2}{3} a+\left(a_{1}-\frac{2}{3} a\right)\left(-\frac{1}{8}\right)^{n-1}$.
Therefore, when $n \rightarrow \infty$, $a_{n} \rightarrow \frac{2}{3} a$, which means point $P_{n}$ approaches the point that divides $A B$ in the ratio $1: 2$.