Given the problem, we need to find all positive integers $a, b, c$ such that each of the expressions $ab + 1$, $bc + 1$, and $ca + 1$ are factorials of positive integers. Let's denote these factorials as follows:

$$ab + 1 = x!$$
$$bc + 1 = y!$$
$$ca + 1 = z!$$

where $x, y, z$ are positive integers.

Since $ab + 1$, $bc + 1$, and $ca + 1$ are all factorials, we can infer:

$$ab = x! - 1$$
$$bc = y! - 1$$
$$ca = z! - 1$$

Without loss of generality, we begin by exploring simple cases. Consider the scenario where at least one of $b$ or $c$ is 1. Assume $b = 1$. This simplifies our equations to:

$$a(1) + 1 = x! \implies a = x! - 1$$
$$1c + 1 = z! \implies c = z! - 1$$

For the second equation:
$$ca + 1 = z!$$
Substitute $a = x! - 1$ into the equation:
$$(z! - 1)(x! - 1) + 1 = y!$$

Given that $a, b, c$ are permutations, we can further explore the symmetrical nature by setting $b = 1$ and $c = 1$, or $a = 1$ and $c = 1$, to find viable solutions.

Exploring these equations, assume $b = 1$ and $c = 1$:

For $ab + 1 = x!$, we have:
$$a(1) + 1 = x! \implies a = x! - 1$$
and for $bc + 1 = y!$:
$$1(1) + 1 = 2 = y!\text{ if }y = 2$$

Thus, $b = 1$, $c = 1$, $a = k! - 1$, where $k$ is a positive integer such that $k > 1$, satisfies the conditions. With these, all the expressions are valid factorials:

- $ab + 1 = (k! - 1) \cdot 1 + 1 = k!$
- $bc + 1 = 1 \cdot 1 + 1 = 2\!,$ for $y=2$
- $ca + 1 = 1 \cdot (k! - 1) + 1 = k!$

Therefore, the solution is:

$$\boxed{(k! - 1, 1, 1)} \text{ (and its permutations), where } k \in \mathbb{N}_{>1}$$