9. As shown in Figure 2, in the polyhedron ABCDEF, it is known that quadrilateral $ABCD$ is a square with side length 3, $EF$ $/ / AB, EF=\frac{3}{2}$. If the volume of the polyhedron is $\frac{15}{2}$, then the distance between $EF$ and $AC$ is
9.2 .

Take the midpoints of $A B$ and $C D$ as $M$ and $N$, and connect $F M$, $F N$, and $M N$.

Since $E F / / A M$ and $E F = \frac{3}{2} = A M$, we know that the polyhedron $A D E - M N F$ is a triangular prism.
Let the distance between $E F$ and $A C$ be $h$. From
$$
V_{A D E-M N F} + V_{F-B C M M} = \frac{15}{2} \text {, }
$$

we get $\frac{1}{2} \cdot h \cdot \frac{9}{2} + \frac{1}{3} \cdot h \cdot \frac{9}{2} = \frac{15}{2}$.
Solving for $h$, we get $h = 2$.