已知曲线的方程为 $y = ax + b + \frac{c}{x^2 - 1}$，其中 $a$、$b$ 和 $c$ 是实常数。已知该曲线在 $x = 2$ 处与 $x$ 轴相交。曲线在点 $(0, 3)$ 处的法线与 $x$ 轴相交于 $x = 7.5$。求 $a$、$b$ 和 $c$ 的值。

1. Form an equation using the coordinates of the point on the normal.

Since the normal is drawn to the curve at the point $(0, 3)$, the point $(0, 3)$ must lie on the curve. Substitute $x = 0$ and $y = 3$ into the curve equation:
$3 = a(0) + b + \frac{c}{0^2 - 1}$
$3 = b - c \implies c = b - 3$

2. Form an equation using the $x$-intercept of the curve.

The curve crosses the $x$-axis at $x = 2$, meaning it passes through $(2, 0)$. Substitute $x = 2$ and $y = 0$ into the curve equation:
$0 = a(2) + b + \frac{c}{2^2 - 1}$
$0 = 2a + b + \frac{c}{3}$
$6a + 3b + c = 0$

3. Calculate the gradient of the normal and deduce the gradient of the tangent at $x = 0$.

The normal passes through the points $(0, 3)$ and $(7.5, 0)$.
$\text{Gradient of normal} = \frac{0 - 3}{7.5 - 0} = -\frac{3}{7.5} = -\frac{2}{5}$
Let $m$ be the gradient of the tangent at $x = 0$. Since the tangent and normal are perpendicular:
$m \times \left(-\frac{2}{5}\right) = -1 \implies m = \frac{5}{2}$

4. Find $\frac{dy}{dx}$ and use the gradient of the tangent to find $a$.

Differentiate the curve equation with respect to $x$:
$y = ax + b + c(x^2 - 1)^{-1}$
$\frac{dy}{dx} = a - c(x^2 - 1)^{-2}(2x) = a - \frac{2cx}{(x^2 - 1)^2}$
At $x = 0$, the gradient of the tangent is $\frac{dy}{dx} = m$:
$a - \frac{2c(0)}{(0^2 - 1)^2} = \frac{5}{2}$
$a = \frac{5}{2}$

5. Solve the simultaneous equations to find $b$ and $c$.

Substitute $a = \frac{5}{2}$ and $c = b - 3$ into the equation $6a + 3b + c = 0$:
$6\left(\frac{5}{2}\right) + 3b + (b - 3) = 0$
$15 + 4b - 3 = 0$
$12 + 4b = 0 \implies 4b = -12 \implies b = -3$
Substitute $b = -3$ into $c = b - 3$:
$c = -3 - 3 = -6$