Let's determine the greatest common divisor of all numbers that can be written in the form $(a-b)(c-d)(a-c)(b-d)(a-d)(b-c)$, where $a, b, c, d$ are arbitrary natural numbers.
We denote the product by $N$. The numbers $a, b, c, d$ can be of two types in terms of divisibility by 2: $2k$ or $2k+1$. Among these four numbers, at least two are of the same parity, and then the other two are also of the same type. But the difference between numbers of the same type is even, so in this case, $N$ has two even factors. The others are odd. In the other case, at least three of the numbers $a, b, c, d$ are of the same type, and thus in this case $N$ is divisible by 8, because 3 differences are even. Therefore, $N$ is always divisible by 4. In terms of divisibility by 3, the numbers can be of the following forms: $3k-1, 3k, 3k+1$. For these three possibilities, we have 4 numbers, so at least two numbers are of the same form, and then their difference is divisible by 3.

For example, $(3a-1)-(3b-1)=3(a-b)$. $N$ is always divisible by 3. However, $N$ is generally not divisible by 8, or 9, $m$ or any other prime number greater than 3. The greatest common divisor of the numbers $N$ is $4 \cdot 3=12$.