<s>[INST] $3$ primitive pythagorean triples from 6 integers. Do there exist $3$ different primitive Pythagorean triples $(a,d,w), (a,b,z)$ and $(c,d,z)$? 
Explicitly, we want $6$ different integers $a,b,c,d,w,z$ such that...
(1) $a^2 + d^2 = w^2$
(2) $a^2 + b^2 = c^2 + d^2 = z^2$
 [/INST] This is surprisingly simple given a function derived from Euclid's formula:
$$A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2$$
To find a triple matching side-C,
$$C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor$$
The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$, for example
$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$
$$33^2+56^2=63^2+16^2=65^2$$
The larger the value of $C$ the greater the chances there are more solutions, for example $C=1105$:
$f(24,23)=(47,1104,1105)\\
f(31,12)=(817,744,1105)\\ 
f(32,9)=(943,576,1105)\\
f(33,4)=(1073,264,1105)$
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