已知双曲线的方程为 $a(x-2)^2 - b(y+1)^2 = c$，其中 $a$、$b$ 和 $c$ 是正整数。双曲线经过点 $(7,10)$。直线 $y = x - 1$ 与该双曲线相交于 $x = 1$ 处。求一组可能满足条件的 $a$、$b$ 和 $c$ 的值。

1. Find the coordinates of the point of intersection with the line.

When $x = 1$, the $y$-coordinate on the line $y = x - 1$ is:
$y = 1 - 1 = 0$
So, the hyperbola passes through the point $(1, 0)$.

2. Set up a system of linear equations using the known points on the hyperbola.

Substitute $(7, 10)$ into the hyperbola's equation:
$a(7-2)^2 - b(10+1)^2 = c$
$25a - 121b = c \quad \text{--- (1)}$

Substitute $(1, 0)$ into the hyperbola's equation:
$a(1-2)^2 - b(0+1)^2 = c$
$a(-1)^2 - b(1)^2 = c$
$a - b = c \quad \text{--- (2)}$

3. Solve the system to express $a$ and $c$ in terms of $b$.

Equating (1) and (2):
$25a - 121b = a - b$
$24a = 120b \implies a = 5b$

Substitute $a = 5b$ into equation (2):
$5b - b = c \implies c = 4b$

4. Find a possible set of positive integer values for $a$, $b$, and $c$.

Since $a$, $b$, and $c$ are positive integers, we can choose the smallest positive integer $b = 1$.
Then:
$a = 5(1) = 5$
$c = 4(1) = 4$