Three. (25 points) Given the quadratic function $y=x^{2}+b x-c$ whose graph passes through two points $P(1, a)$ and $Q(2,10 a)$.
(1) If $a$, $b$, and $c$ are all integers, and $c<b<8 a$, find the values of $a$, $b$, and $c$;
(2) Let the graph of the quadratic function $y=x^{2}+b x-c$ intersect the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If the roots of the equation $x^{2}+b x-c=0$ are both integers, find the area of $\triangle A B C$.
Three, given that points $P(1, a)$ and $Q(2, 10a)$ are on the graph of the quadratic function $y = x^2 + bx - c$, we have
$$
1 + b - c = a, \quad 4 + 2b - c = 10a.
$$

Solving these equations, we get $b = 9a - 3$ and $c = 8a - 2$.
(1) From $c < b < 8a$, we know
$$
8a - 2 < 9a - 3 < 8a \Rightarrow 1 < a < 3.
$$
Since $a$ is an integer, we have
$$
a = 2, \quad b = 9a - 3 = 15, \quad c = 8a - 2 = 14.
$$
(2) Let $m$ and $n$ be the two integer roots of the equation, with $m \leq n$. By the relationship between roots and coefficients, we have
$$
m + n = -b = 3 - 9a, \quad mn = -c = 2 - 8a.
$$

Eliminating $a$, we get $9mn - 8(m + n) = -6$.
Multiplying both sides by 9 and factoring, we get
$$
(9m - 8)(9n - 8) = 10.
$$

Thus, $(9m - 8, 9n - 8)$
$$
= (1, 10), (2, 5), (-10, -1), (-5, -2).
$$

Solving these, we get $(m, n)$
$$
= (1, 2), \left(\frac{10}{9}, \frac{13}{9}\right), \left(-\frac{2}{9}, \frac{7}{9}\right), \left(\frac{1}{3}, \frac{2}{3}\right).
$$

Since $m$ and $n$ are integers, we have $(m, n) = (1, 2)$.
Therefore, $b = -(m + n) = -3$,
$$
c = -mn = -2.
$$

Thus, the quadratic function is
$$
y = x^2 - 3x + 2.
$$

It is easy to find that $A(1, 0)$, $B(2, 0)$, and $C(0, 2)$.
So, $S_{\triangle ABC} = \frac{1}{2} \times (2 - 1) \times 2 = 1$.