Three. (25 points) Given the system of equations about $x$ and $y$:
$$
\left\{\begin{array}{l}
k x^{2} - a y + k - 3 b c = 0, \\
y = (k-3) x
\end{array}\right.
$$

has only one integer solution. Here, $k, a, b, c$ are all integers, and $a > 0$, $a, b, c$ satisfy $a^{2} - a - b c = -1$, $b + c = 2$. Find:
(1) The value of $a$;
(2) The value of $k$ and the corresponding integer solution $(x, y)$.
(1) From $b+c=2, bc=a^{2}-a+1$ we know that $b, c$ are the two real roots of the equation $x^{2}-2x+a^{2}-a+1=0$, then we have
$$
\Delta=(-2)^{2}-4\left(a^{2}-a+1\right) \geqslant 0 \text {. }
$$

Therefore, $a^{2}-a \leqslant 0$.
Since $a>0$, then $a \leqslant 1$. Also, $a$ is a positive integer, so, $a=1$.
(2) When $a=1$, $bc=1$, the original equation can be transformed into
$$
\left\{\begin{array}{l}
k x^{2}-y+k-3=0, \\
y=(k-3) x .
\end{array}\right.
$$
(i) When $k=0$, the system of equations becomes $\left\{\begin{array}{l}y+3=0 \\ y=-3 x\end{array}\right.$.

So, the system of equations has a unique integer solution $x=1, y=-3$.
(ii) When $k \neq 0$, the system of equations can be transformed into
$$
k x^{2}-(k-3) x+k-3=0 \text {. }
$$

Since the system of equations has integer solutions, then equation (2) must have real roots. So,
$$
\begin{array}{l}
\Delta=(k-3)^{2}-4 k(k-3) \\
=-3(k-3)(k+1) \geqslant 0 .
\end{array}
$$

Solving gives $-1 \leqslant k \leqslant 3$ and $k \neq 0$.
Since $k$ is an integer, so, $k=-1,1,2,3$.
Substituting into the system of equations (1) and verifying, we get
$$
\begin{array}{l}
k=-1 \text {, } x=2, y=-8 ; \\
k=2 \text {, } x=-1, y=1 ; \\
k=3 \text {, } x=0, y=0 .
\end{array}
$$