Three, (13 points) As shown in Figure 4, two planes $m \parallel n$, line segment $AD$ intersects planes $m$ and $n$ at points $B$ and $C$, respectively. Another line through point $A$ intersects planes $m$ and $n$ at points $M$ and $P$, respectively. Another line through point $D$ intersects planes $m$ and $n$ at points $N$ and $Q$, respectively. Given $\frac{S_{\triangle BMN}}{S_{\triangle CPQ}}=\frac{3}{2}$. Find the minimum value of $\frac{AD}{CD}$.
Three, it is known that $B M / / C P, B N / / C Q$. Then $\sin \angle M B N=\sin \angle P C Q$,

and
$$
\begin{array}{l}
\frac{B M}{C P}=\frac{A B}{A C}, \frac{B N}{C Q}=\frac{B D}{C D}. \\
\text { Also } S_{\triangle B M N}=\frac{1}{2} B M \cdot B N \sin \angle M B N, \\
S_{\triangle C P Q}=\frac{1}{2} C P \cdot C Q \sin \angle P C Q,
\end{array}
$$

Combining $\frac{S_{\triangle B M N}}{S_{\triangle C P Q}}=\frac{3}{2}$, we get $\frac{A B}{A C} \cdot \frac{B D}{C D}=\frac{3}{2}$.
$$
\text { Let } \frac{A C}{A B}=a, \frac{B D}{C D}=b .
$$

Then $2 b=3 a$, and $b>a>1$.
$$
\begin{array}{l}
\text { Note that } \frac{B C}{A C}=a-1, A B=\frac{1}{a-1} B C . \\
\text { Therefore } \frac{A D}{C D}=\frac{A B+B C+C D}{C D}=1+\frac{a}{a-1} \cdot \frac{B C}{C D} \\
=1+\frac{a}{a-1} \cdot \frac{B D-C D}{C D}=1+\frac{a}{a-1}(b-1) \\
=1+\frac{a}{a-1}\left(\frac{3}{2} a-1\right) \\
=\frac{3}{2}\left[(a-1)+\frac{1}{3(a-1)}\right]+3 \\
\geqslant \frac{3}{2} \times 2 \sqrt{\frac{1}{3}}+3=3+\sqrt{3} .
\end{array}
$$

Therefore, the minimum value of $\frac{A D}{C D}$ is $3+\sqrt{3}$.