Three primitive Pythagorean triples with the same c

We produce an answer with hypotenuse $c=5\cdot 13\cdot 17$. So we need to find three relatively prime pairs $\{m,n\}$ of opposite parity such that $m^2+n^2=5\cdot 13\cdot 17$.

It is simplest to calculate using complex numbers, although the same goal can be achieved by using the Brahmagupta Identity. Note that $c$ factors over the Gaussian integers as $(2+i)(2-i)(3+2i)(3-2i)(4+i)(4-i)$.

First one: Note that $(2+i)(3+2i)(4+i)=9+32i$. That gives $m=9$, $n=32$.

Second one: Do the same calculation using $(2-i)(3+2i)(4+i)$.

Third one: Same calculation, using $(2+i)(3-2i)(4+i)$.

Remark: There is a fourth, using $(2+i)(3+i)(4-i)$. For an example with $8$ instead of $4$, we can play the same game with $5\cdot 13\cdot 17\cdot 29$.

This is a list of the first Pythagorean triples. $1105$ actually has four primitive breakdowns as $a^2+b^2$, and my skimming did not find any earlier ones with three.

$$\begin{align} 1105^2 &=47^2+1104^2\\ &=264^2+1073^2\\ &=576^2+943^2\\ &=744^2+817^2 \end{align}$$

(Maybe @AndreNicolas's answer can be explored more to prove that the number of breakdowns for any $c$ that is the hypotenuse of a Pythagorean triple is a power of $2$.)