3. As shown in Figure $1, \angle A O B=30^{\circ}$, within $\angle A O B$ there is a fixed point $P$, and $O P$ $=10$, on $O A$ there is a point $Q$, and on $O B$ there is a fixed point $R$. To make the perimeter of $\_P Q R$ the smallest, the minimum perimeter is
3.10 .

As shown in Figure 5, construct the symmetric points $P_{1}, P_{2}$ of $P$ with respect to the sides $O A, O B$ of $\angle A O B$. Connect $P_{1} P_{2}$, intersecting $O A, O B$ at $Q, R$. Connect $P Q, P R$. It is easy to see that
$$
P Q=P_{1} Q, P R=P_{2} R,
$$

which implies
$$
\begin{aligned}
l_{\triangle P Q N} & =P_{1} Q+Q R+P_{2} R \\
& =P_{1} P_{2} .
\end{aligned}
$$

Thus, $\triangle P Q R$ is the triangle with the minimum perimeter.
Given that $O P=O P_{1}=O P_{2}=10$,
$$
\angle P O Q=\angle P_{1} O Q, \angle P O R=\angle P_{2} O R,
$$

we have
$$
\begin{aligned}
\angle P_{1} O P_{2} & =2 \angle P O Q+2 \angle P O R \\
& =2(\angle P O Q+\angle P O R)=2 \angle A O B=60^{\circ} .
\end{aligned}
$$

Therefore, $\triangle P_{1} O P_{2}$ is an equilateral triangle,
Thus, $P_{1} P_{2}=P_{1} O=10$.
Hence, the minimum perimeter of $\triangle P Q R$ is 10.