15. (12 points) As shown in Figure 1, plane $m \parallel$ plane $n$, 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, and another line through point $D$ intersects planes $m$ and $n$ at points $N$ and $Q$ respectively. Given that $\frac{S_{\triangle BMN}}{S_{\triangle CPQ}}=\frac{3}{2}$. Find the minimum value of $\frac{AD}{CD}$.
Three, 15. From the fact that plane $m //$ plane $n$, we know $B M // C P, B N // C Q$.
Therefore, $\sin \angle M B N = \sin \angle P C Q$, and
$\frac{B M}{C P} = \frac{A B}{A C}, \frac{B N}{C Q} = \frac{B D}{C D}$.
Also, $S_{\triangle B M N} = \frac{1}{2} B M \cdot B N \sin \angle M B N$,
$S_{\triangle C P Q} = \frac{1}{2} C P \cdot C Q \sin \angle P C Q$,
and $\frac{S_{\triangle B M N}}{S_{\triangle C P Q}} = \frac{3}{2}$, we get $\frac{A B}{A C} \cdot \frac{B D}{C D} = \frac{3}{2}$.
Let $\frac{A C}{A B} = a, \frac{B D}{C D} = b$. Then $2 b = 3 a$, and $b > a > 1$.
$$
\begin{array}{l}
\frac{B C}{A B} = \frac{A C - A B}{A B} = a - 1, A B = \frac{1}{a - 1} B C . \\
\frac{A D}{C D} = \frac{A B + B C + C D}{C D} = 1 + \frac{a}{a - 1} \cdot \frac{B C}{C D} \\
= 1 + \frac{a}{a - 1} \cdot \frac{B D - C D}{C D} = 1 + \frac{a}{a - 1}(b - 1) \\
= 1 + \frac{a}{a - 1}\left(\frac{3}{2} a - 1\right) = \frac{3 a^{2} - 2}{2(a - 1)} \\
= \frac{3}{2}(a + 1) + \frac{1}{2(a - 1)} \\
= \frac{3}{2}\left[(a - 1) + \frac{1}{3(a - 1)}\right] + 3 \\
\geqslant \frac{3}{2} \cdot 2 \sqrt{\frac{1}{3}(a - 1) \frac{1}{a - 1}} + 3 \\
= 3 + \sqrt{3} .
\end{array}
$$

Clearly, when $a - 1 = \frac{\sqrt{3}}{3}$, i.e., $a = 1 + \frac{\sqrt{3}}{3}$, $\frac{A D}{C D} = 3 + \sqrt{3}$. Therefore, the minimum value of $\frac{A D}{C D}$ is $3 + \sqrt{3}$.