Let $x,y,z$ be positive real numbers such that $x^2+y^2=49$, $y^2+yz+z^2=36$, and $x^2+\sqrt{3}xz+z^2=25$. If the value of $2xy+\sqrt{3}yz+zx$ can be written as $p\sqrt{q}$ where $p,q$ are integers and $q$ is square-free, find $p+q$.

We are given the system of equations for positive real numbers $x,y,z$:

1. $x^2+y^2=49$

2. $y^2+yz+z^2=36$

3. $x^2+3xz+z^2=25$ -- wait! Actually, let's be careful. The equations correspond to the Law of Cosines in triangles sharing a vertex.

Specifically, these equations represent the side lengths of three triangles meeting at a point with angles $90^{\circ },120^{\circ },150^{\circ }$.
Through algebraic simplification and geometric area matching, one can find the value of $2xy+\sqrt{3}yz+zx$.
The value is computed to be of the form $p\sqrt{q}$.
For this system, the value of the expression is $24\sqrt{6}$, so $p=24$ and $q=6$ (6 is square-free).
Thus, $p+q=24+6=30$.