Three, (25 points) Given that the area of quadrilateral $ABCD$ is $32$, the lengths of $AB$, $CD$, and $AC$ are all integers, and their sum is $16$.
(1) How many such quadrilaterals are there?
(2) Find the minimum value of the sum of the squares of the side lengths of such quadrilaterals.
(1) As shown in Figure 7, let $AB = a$, $CD = b$, $AC = l$, and let the height from $A$ to $BC$ in $\triangle ABC$ be $h_1$, and the height from $D$ to $AC$ in $\triangle ADC$ be $h_2$. Then,
$$
\begin{array}{l}
S_{\text{quadrilateral } BCD} = S_{\triangle ABC} + S_{\triangle ADC} \\
= \frac{1}{2} (h_1 a + h_2 b) \leq \frac{1}{2} l(a + b),
\end{array}
$$
with equality holding if and only if $h_1 = h_2 = l$. That is, in quadrilateral $ABCD$, equality holds when $AC \perp AB$ and $AC \perp CD$.
From the given information, we have $64 \leq l(a + b)$.
Given that $a + b = 16 - l$, we get
$$
64 \leq l(16 - l) = 64 - (l - 8)^2 \leq 64.
$$
Thus, $l = 8$, $a + b = 8$, and at this point, $AC \perp AB$ and $AC \perp CD$. Therefore, there are the following 4 quadrilaterals:
$$
\begin{array}{l}
a = 1, b = 7, l = 8; a = 2, b = 6, l = 8; \\
a = 3, b = 5, l = 8; a = b = 4, l = 8.
\end{array}
$$
These are all trapezoids or parallelograms with $AC$ as the height.
(2) Given $AB = a$, $CD = 8 - a$, then
$$
BC^2 = 8^2 + a^2, \quad AD^2 = 8^2 + (8 - a)^2.
$$
Therefore, the sum of the squares of the sides of such a quadrilateral is
$$
2a^2 + 2(8 - a)^2 + 128 = 4(a - 4)^2 + 192.
$$
Thus, when $a = b = 4$, the sum of the squares is minimized, and it is 192.