9. (14 points) As shown in Figure 2, the corridor is 3 m wide, the angle between the corridors is 120°, the ground is level, and the ends of the corridor are sufficiently long. Question: What is the maximum length of a horizontal rod (neglecting its thickness) that can pass through the corridor?
II. 9. As shown in Figure 4, draw any horizontal line through the inner vertex $P$ of the corridor corner, intersecting the outer sides of the corridor at points $A$ and $B$. The length of a wooden rod that can pass through the corridor in a horizontal position is less than or equal to $AB$.
Let $\angle B A Q=\alpha$. Then
$$
\begin{array}{l}
\angle A B Q=60^{\circ}-\alpha, \\
A B=A P+P B=\frac{3}{\sin \alpha}+\frac{3}{\sin \left(60^{\circ}-\alpha\right)} .
\end{array}
$$

As $\alpha$ varies, the minimum value of the above expression is the maximum length of the wooden rod that can pass through the corridor in a horizontal position.
By the arithmetic mean inequality and product-to-sum formulas, we get
$$
\begin{array}{l}
A B \geqslant 6 \sqrt{\frac{1}{\sin \alpha \cdot \sin \left(60^{\circ}-\alpha\right)}} \\
=6 \sqrt{\frac{1}{\frac{1}{2}\left[\cos \left(2 \alpha-60^{\circ}\right)-\cos 60^{\circ}\right]}} \\
\geqslant 6 \sqrt{\frac{2}{1-\frac{1}{2}}}=12,
\end{array}
$$

and when $\alpha=30^{\circ}$, $A B=12$.
Therefore, the maximum length of a wooden rod that can pass through the corridor in a horizontal position is $12 \mathrm{~m}$.