The updated answer is:
$r = 8$
$b = 4$
$y = 3$
I'm making this answer a community wiki, since I made a mistake at the end of my original answer and it was Van.Graaf who delivered the final punch.
Reasoning:
Let's consider all the tri-colored triangles (e.g. one vertex red, one vertex blue, one vertex yellow) and apply triangle inequalities. The sum of lengths of all the red-blue segments must be no greater than the sum of all lengths of all the red-yellow segments and all the blue-yellow ones.
What is the sum of lengths of all the red-blue segments? It is $51y$ (each red-blue pair is counted $y$ times and the sum of pair lengths is $51$). Of red-yellow segments — $39b$, of blue-yellow — $r$. Therefore, the inequalities are:
$51y \le 39b + r$
$39b \le 51y + r$
$r \le 51y + 39b$
$r + b + y = 15$
The only triple $(r, b, y)$ which satisfies the above set of constraints is $(8, 4, 3)$.
Here's the space of possible solutions, with blue on the $x$ axis and yellow on the $y$ axis:
The graph in Desmos: https://www.desmos.com/calculator/per0tkuthz
A set of points with this property can be realized
in something of a trivial way: construct a triangle $RBY$ with sides $RB = 51/32$, $RY = 39/24$, and $BY = 1/12$. Note that these lengths satisfy the triangle inequalities. Then place all eight red points at $R$, all four blue points at $B$, and all three yellow points at $Y$. A scale drawing of this configuration is below; the points used are $(0,0)$, $(0,1/12)$, and $(\sqrt{5247935}/1536, 991/1536)$.
