Three. (25 points) Let $a$ be a prime number, $b$ and $c$ be positive integers, and satisfy
$$
\left\{\begin{array}{l}
9(2 a+2 b-c)^{2}=509(4 a+1022 b-511 c), \\
b-c=2 .
\end{array}\right.
$$

Find the value of $a(b+c)$.
Three, Equation (1) is
$$
\left(\frac{6 a+6 b-3 c}{509}\right)^{2}=\frac{4 a+1022 b-511 c}{509} \text {. }
$$

Let $m=\frac{6 a+6 b-3 c}{509}$,
$$
n=\frac{4 a+1022 b-511 c}{509} \text {. }
$$

Then $2 b-c=\frac{509 m-6 a}{3}=\frac{509 n-4 a}{511}$.
Thus, $3 n-511 m+6 a=0$.
Since $n=m^{2}$, we have
$$
3 m^{2}-511 m+6 a=0 \text {. }
$$

From Equation (1), we know that $(2 a+2 b-c)^{2}$ is divisible by 509, and since 509 is a prime number, $2 a+2 b-c$ is divisible by 509. Therefore, $m$ is an integer, meaning the quadratic equation (4) in $m$ has integer roots. Hence, its discriminant $\Delta=511^{2}-72 a$ is a perfect square.

Assume $\Delta=511^{2}-72 a=t^{2}$ (where $t$ is a natural number). Then $72 a=511^{2}-t^{2}=(511+t)(511-t)$.

Since $511+t$ and $511-t$ have the same parity, and $511+t \geqslant 511$, there are only the following possible cases:
(1) $\left\{\begin{array}{l}511+t=36 a, \\ 511-t=2 .\end{array}\right.$
Adding the two equations gives $36 a+2=1022$, which has no integer solution;
(2) $\left\{\begin{array}{l}511+t=18 a, \\ 511-t=4 .\end{array}\right.$
Adding the two equations gives $18 a+4=1022$, which has no integer solution;
(3) $\left\{\begin{array}{l}511+t=12 a, \\ 511-t=6 .\end{array}\right.$
Adding the two equations gives $12 a+6=1022$, which has no integer solution;
Adding the two equations gives $6 a+12=1022$, which has no integer solution;
Adding the two equations gives $4 a+18=1022$, solving for $a$ gives $a=251$;
(6) $\left\{\begin{array}{l}511+t=2 a \\ 511-t=36\end{array}\right.$.
Adding the two equations gives $2 a+36=1022$, solving for $a$ gives $a=493$, but $493=17 \times 29$ is not a prime number, so it is discarded.
In summary, $a=251$.
At this point, the solution to Equation (4) is
$m=3$ or $m=\frac{502}{3}$ (discarded).
Substituting $a=251, m=3$ into Equation (3) gives
$2 b-c=\frac{509 \times 3-6 \times 251}{3}=7$,

i.e., $c=2 b-7$.
Substituting into Equation (2) gives $b-(2 b-7)=2$, so $b=5, c=3$.
Therefore, $a(b+c)=251 \times(5+3)=2008$.
(Provided by Xu Shenglin)