Is there a formula for this specific pattern of Pythagorean Triangles sharing an area?

A subset of your 11 data points comprise another infinite family that solves,

$$M_1 N_1 (M_1^2-N_1^2)=M_2 N_2 (M_2^2-N_2^2)=M_3 N_3 (M_3^2-N_3^2)$$

distinct from the one given by Enrique Zeleny. You didn't notice that,

$(10,4),(12,\color{red}2),(15,\color{red}1)$
$(20,6),(21,5),(28,2)$
$(24,14),(35,\color{red}3),(40,\color{red}2)$
$(42,20),(55,7),(66,4)$
$(44,30),(70,\color{red}4),(77,\color{red}3)$
$(56,30),(78,8),(91,5)$
$(65,33),(88,10),(104,6)$
$(70,52),(117,\color{red}5),(126,\color{red}4)$

The high-lighted numbers then give us a clue there is a second family. Thus, to summarize,

1st family (Zeleny's). Special case: $M_1 = M_2 = N_3$

$M_1 = r^2 + rs + s^2;\quad N_1 = r^2 - s^2$

$M_2 = r^2 + rs + s^2;\quad N_2 = 2rs + s^2$

$M_3 = r^2 + 2rs;\quad\quad\; N_3 = r^2 + rs + s^2$

2nd family. Special case: $M_1+N_1 = M_2+N_2 = M_3-N_3$

$M_1 = (u + v) (2u + 3 v);\quad N_1 = v (u + 3 v)$

$M_2 = (u + 2 v) (u + 3 v);\quad N_2 = u(u + v)$

$M_3 = (u + 2 v) (2u + 3 v);\quad N_3 = uv$

where $M_1+N_1 = M_2+N_2 = M_3-N_3 = 2 (u^2 + 3 u v + 3 v^2)$.

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