12. In the park, there are two rivers $O M$ and $O N$ converging at point $O$ (as shown in Figure 6, $\angle M O N=60^{\circ}$. On the peninsula formed by the two rivers, there is an ancient site $P$. It is planned to build a small bridge $Q$ and $R$ on each of the two rivers, and to construct three small roads to connect the two bridges $Q$, $R$, and the ancient site $P$. If the distance from the ancient site $P$ to the two rivers is $50 \sqrt{3} \mathrm{~m}$ each, then the minimum value of the sum of the lengths of the three small roads is $\qquad$ m.
12.300.

As shown in Figure 10, let the symmetric points of point $P$ with respect to $OM$ and $ON$ be $P_{1}$ and $P_{2}$, respectively. Connect $P_{1}P_{2}$, intersecting $OM$ and $ON$ at points $Q$ and $R$. By symmetry, we have
$$
\begin{array}{l}
PQ = P_{1}Q, \\
PR = P_{2}R, \\
PQ + QR + RP \\
= P_{1}Q + QR + RP_{2} = P_{1}P_{2}.
\end{array}
$$

Now we prove that $PQ + QR + RP$ is the minimum value sought.
Take any point $Q'$ on $OM$ (not coinciding with point $Q$), and any point $R'$ on $ON$ (not coinciding with point $R$). Connect $P_{1}Q'$, $Q'R'$, and $R'P_{2}$. By symmetry, we have
$$
PQ' = P_{1}Q', \quad PR' = P_{2}R'.
$$

Then $PQ' + Q'R' + R'P = P_{1}Q' + Q'R' + R'P_{2}$
$$
> P_{1}P_{2} = PQ + QR + RP.
$$

Therefore, $PQ + QR + RP$ is the minimum value sought.
Given in $\triangle PP_{1}P_{2}$,
$$
\begin{array}{l}
PP_{1} = PP_{2} = 100\sqrt{3}, \\
\angle P_{1}PP_{2} = 120^{\circ}, \\
\angle P_{1} = \angle P_{2} = 30^{\circ},
\end{array}
$$

Construct $PK \perp P_{1}P_{2}$, with the foot of the perpendicular at $K$. Then in the right triangle $\triangle PP_{1}K$,
$$
\begin{array}{l}
PK = \frac{1}{2} PP_{1} = 50\sqrt{3}, \\
P_{1}K = \sqrt{PP_{1}^2 - PK^2} \\
= \sqrt{(100\sqrt{3})^2 - (50\sqrt{3})^2} = 150.
\end{array}
$$

Thus, $P_{1}P_{2} = 2P_{1}K = 300$.
Therefore, the minimum value of the sum of the lengths of the three paths is
$$
PQ + QR + RP = 300 \text{ m}.
$$