Solution:
The equation is equivalent to $x \in (1,2)$ and $a x + 1 = (x-1)(2-x)$, which can be written as $x^2 + (a-3)x + 3 = 0$. Therefore we have to find the values of $a$ such that the equation
$$f(x) = x^2 + (a-3)x + 3 = 0$$
has exactly one root in the interval $(1,2)$. This is possible exactly in the following four cases:

Case 1. $f(1) f(2) < 0$, which is equivalent to $a \in \left(-1, -\frac{1}{2}\right)$.

Case 2. $f(1) = 0$, i.e. $a = -1$. Then $x_1 = 1$ and $x_2 = 3$, which shows that $a = -1$ is not a solution.

Case 3. $f(2) = 0$, i.e. $a = -\frac{1}{2}$. Then $x_1 = 2$ and $x_2 = \frac{3}{2}$, which shows that $a = -\frac{1}{2}$ is a solution.

Case 4. $D = 0$, i.e. $(a-3)^2 - 12 = a^2 - 6a - 3 = 0$, whence $a = 3 \pm 2\sqrt{3}$. For $a = 3 + 2\sqrt{3}$ we have $x_1 = x_2 = -\sqrt{3}$, i.e. $a = 3 + 2\sqrt{3}$ is not a solution. For $a = 3 - 2\sqrt{3}$ we get $x_1 = x_2 = \sqrt{3} \in (1,2)$, i.e. $a = 3 - 2\sqrt{3}$ is a solution.

Finally, $a \in \left(-1, -\frac{1}{2}\right] \cup \{3 - 2\sqrt{3}\}$.