Three, (25 points) Given three natural numbers $a, b, c$ where at least $a$ is a prime number, and they satisfy
$$
\left\{\begin{array}{l}
(4 a+2 b-4 c)^{2}=443(2 a-442 b+884 c), \\
\sqrt{4 a+2 b-4 c+886}-\sqrt{42 b-2 a+2 c-443}=\sqrt{443} .
\end{array}\right.
$$

Find the value of $a b c$.
Let $x=\frac{4 a+2 b-4 c}{443}, y=\frac{2 a-442 b+884 c}{443}$. Then
$4 a+2 b-4 c=443 x$,
$2 a-442 b+884 c=443 y$.
Hence $x^{2}=y$.
From equation (1), we know that $(4 a+2 b-4 c)^{2}$ is divisible by 443. Since 443 is a prime number, 443 divides $(4 a+2 b-4 c)$. Therefore, $x$ is an integer.
From equation (5), we know that $y$ is an integer.
(3) $\times 221+$ (4) gives
$$
4 \times 221 a+2 a=443 \times 221 x+443 y \text {, }
$$

which simplifies to $2 a=221 x+y$.
(5) - (6) gives $x^{2}+221 x-2 a=0$.
Then $\Delta=221^{2}-4 \times 1 \times(-2 a)=221^{2}+8 a$.
Since $x$ is an integer, $\Delta$ must be a perfect square.
Let $\Delta=t^{2}(t \in \mathbf{N})$. Then $t^{2}=221^{2}+8 a$, which implies $(t+221)(t-221)=8 a$.
Since $t+221$ and $t-221$ have the same parity and $t+221>221$, we have $\left\{\begin{array}{l}t+221=4 a, \\ t-221=2\end{array}\right.$ or $\left\{\begin{array}{l}t+221=2 a, \\ t-221=4 .\end{array}\right.$
Solving these, we get $\left\{\begin{array}{l}t=223, \\ a=111\end{array}\right.$ (discard) or $\left\{\begin{array}{l}t=225, \\ a=223 .\end{array}\right.$
Therefore, $x^{2}+221 x-2 \times 223=0$.
Thus, $x=2$ or $x=-223$.
From equation (1), we know that $x$ is even, so $x=2$.
From equation (3), we know that $b=2 c-3$.
Substituting $a=223, b=2 c-3$ into equation (2) gives
$$
c=3, b=3 \text {. }
$$

Therefore, $a b c=223 \times 3 \times 3=2007$.
(Xie Wenxiao, Huanggang High School, Hubei Province, 438000)