Your formula does generate Pythagorean triples but misses most of them and appears to require seeds to work.
I'm not sure what you are generating. You do generate triples where $C-A=2$ in the first column but that can be generated more easily by
$\quad A=4n^2-1\quad B=4n\quad C=4n^2+1.\quad $ The rest of the table shows no pattern that I can see, like a consistent side difference within a set or consistent increment of side values within a set. The following formula generates all primitives and a few that are not but there is a consistent
$C-B=(2n-1)^2\quad$ and $\quad A_{n+1}-A_{n}=2(2n-1).\quad$ It is the formula derive when
$A=(2n-1+k)^2-k^2,\space
B=2(2n-1+k)k,\space
C=(2n-1+k)^2+k^2$
\begin{align*}
A=(2n-1)^2+ \quad &2(2n-1)k\\
B=\hspace{55pt} &2(2n-1)k\quad+2k^2\\
C=(2n-1)^2+ \quad &2(2n-1)k\quad +2k^2
\end{align*}
Here is a sample of what it generates
$$\begin{array}{c|c|c|c|c|}
n & k=1 & k=2 & k=3 & k=4 \\ \hline
Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41 \\ \hline
Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 \\ \hline
Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline
Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 \\ \hline
Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 \\ \hline
\end{array}$$
Your formula generates the first column but nothing with a pattern I can see in the other cells. If you do want to work with areas, there is a list of them here. If you can figure out how to generate this sequence, I can show you how to find all of the
$1,\space 2, \text{ or } 3\space $ triples
that correspond to each area.
Generating the Sets
Given the sets in the question, we can match each triple to one generated by the classical formula: $\left(m^2-n^2,2mn,m^2+n^2\right)$ where $(m,n)=1$, $2\not\mid m-n\gt0$:
$$
\begin{array}{c|c}
\text{set}_0&3,4,5&5,12,13&7,24,25&9,40,41\\\hline
m,n&2,1&3,2&4,3&5,4\\\hline\text{set}_1&15,8,17&33,56,65&51,140,149&69,260,269\\\hline
m,n&4,1&7,4&10,7&13,10\\\hline
\text{set}_2&35,12,37&85,132,157&135,352,377&185,672,697\\\hline
m,n&6,1&11,6&16,11&21,16\\\hline
\text{set}_3 &63,16,65&161,240,289&259,660,709&357,1276,1325\\\hline
m,n&8,1&15,8&22,15&29,22\\\hline
\text{set}_4&99,20,101&261,380,461&423,1064,1145&585,2072,2153\\\hline
m,n&10,1&19,10&28,19&37,28\\\hline
\end{array}
$$
Noticing the pattern in the table above, we get that
$$
\begin{align}
\text{column $j$ of set}_k&=\left(m^2-n^2,2mn,m^2+n^2\right)\\
\text{where }(m,n)&=(1+(2k+1)(j+1),1+(2k+1)j)
\end{align}
$$
Column $0$ is the leftmost column of the table.
To see that $(m,n)=1$, note that $n(j+1)-mj=1$
To see that $2\not\mid m-n\gt0$, note that $m-n=2k+1$
I have added $\text{set}_0$ to the table, following the pattern in the subsequent sets.
These sets do not cover all of the Pythagorean Triples. For example, $(21,20,29)$ is not covered in any of these sets.
Original Answer: Why $\boldsymbol{ab-12r^2=a'b'}$
This answer, and probably many others, shows the following: all relatively prime Pythagorean triples can be written as $\left\{m^2-n^2,2mn,m^2+n^2\right\}$ where $(m,n)=1,\ 2\nmid m-n\gt0$.
The area of a triangle is the inradius times the semi-perimeter. Since a Pythagorean triangle is a right triangle, the area is half the product of the legs. Thus, the inradius is
$$
\begin{align}
\text{inradius}
&=\frac{\text{area}}{\text{semi-perimeter}}\\
&=\frac{mn\left(m^2-n^2\right)}{m^2+mn}\\[9pt]
&=n(m-n)
\end{align}
$$
Suppose that $\{a,b,c\}=\left\{m^2-n^2,2mn,m^2+n^2\right\}$ is a Pythagorean triple. Then
$$
\begin{align}
\overbrace{2mn\left(m^2-n^2\right)}^{\large ab}-12\overbrace{n^2(m-n)^2}^{\large r^2}
&=2m(m+n)n(m-n)-12n(m-n)n(m-n)\\
&=2n(m-n)(m(m+n)-6n(m-n))\\
&=2n(m-n)\left(m^2-5mn+6n^2\right)\\
&=2n(m-n)(m-2n)(m-3n)\\
&=\underbrace{2n(m-2n)\vphantom{\left(n^2\right)}}_{\large e}\underbrace{\left((m-2n)^2-n^2\right)}_{\large d}
\end{align}
$$
where $\{d,e,f\}=\left\{(m-2n)^2-n^2,2n(m-2n),(m-2n)^2+n^2\right\}$ is another Pythagorean triple.
In fact:
$$
\begin{bmatrix}d\\e\\f\end{bmatrix}
=\begin{bmatrix}
-1&-2&2\\
2&1&-2\\
-2&-2&3
\end{bmatrix}
\begin{bmatrix}a\\b\\c\end{bmatrix}
$$
and, inversely,
$$
\begin{bmatrix}
-1&2&2\\
-2&1&2\\
-2&2&3
\end{bmatrix}
\begin{bmatrix}d\\e\\f\end{bmatrix}
=\begin{bmatrix}a\\b\\c\end{bmatrix}
$$
Note that $d,e,f$ have the same parity as $a,b,c$, respectively.
Because
$$
\begin{bmatrix}
-1&-2&2\\
2&1&-2\\
-2&-2&3
\end{bmatrix}^T
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&-1
\end{bmatrix}
\begin{bmatrix}
-1&-2&2\\
2&1&-2\\
-2&-2&3
\end{bmatrix}
=\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&-1
\end{bmatrix}
$$
we have $d^2+e^2-f^2=a^2+b^2-c^2$.
Furthermore, as mentioned above,
$$
\begin{align}
\text{inradius}
&=\frac{\text{area}}{\text{semi-perimeter}}\\[6pt]
&=\frac{ab}{a+b+\sqrt{a^2+b^2}}\\
&=\frac{a+b-\sqrt{a^2+b^2}}2
\end{align}
$$
Best Answer
It should be $$(p^2-q^2)^2+(2pq)^2=(p^2+q^2)^2,$$ where $p$ and $q$ are natural numbers such that $p>q$, $\gcd(p,q)=1$ and $p$ and $q$ have different parity.
Now, you can get all triples: $(d(p^2-q^2),d(2pq),d(p^2+q^2)),$ where $d$ is a natural number.