F(x) = (3x + 1) / ((x - 1)(x - 2)(x + 3))

$$x-1 = 0 \implies x = 1$$ $$x-2 = 0 \implies x = 2$$ $$x+3 = 0 \implies x = -3$$

These are the values of $x$ for which the denominator is zero. Therefore, these values must be excluded from the domain.

The domain of the function is all real numbers except $x=1$, $x=2$, and $x=-3$.
In set-builder notation:

$$\boxed{\left\{x \in \mathbb{R} \mid x \neq -3, x \neq 1, x \neq 2\right\}}$$

In interval notation:

$$\boxed{\left(-\infty, -3\right) \cup \left(-3, 1\right) \cup \left(1, 2\right) \cup \left(2, \infty\right)}$$