Solution:

Denote by $\omega$, $\Gamma_{1}$, $\Gamma_{2}$, and $\Gamma_{3}$ the circumcenters of triangles $A B C$, $B O C$, $C O A$, and $A O B$, respectively. An inversion about $\omega$ interchanges $\Gamma_{1}$ and line $B C$, $\Gamma_{2}$ and line $C A$, and $\Gamma_{3}$ and line $A B$. This inversion also preserves tangency between generalized circles, so the image of $\Gamma$ is a circle tangent to $A B$, $B C$, and $C A$. It is the incircle of $A B C$ because it is closer to $O$ than these lines and $A B C$ is acute.

Now we run a few standard calculations. Where $s$, $r$, and $R$ denote the semiperimeter, inradius, and circumradius of $A B C$, respectively, we have the following:
$$\begin{aligned} & [A B C]=\sqrt{s(s-a)(s-b)(s-c)}=\frac{15 \sqrt{7}}{4} \\ & r=[A B C] / s=\sqrt{7} / 2 \\ & R=\frac{a b c}{4[A B C]}=\frac{8}{\sqrt{7}} \\ & O I^{2}=R(R-2 r)=\frac{8}{7} \end{aligned}$$
Let $O I$ intersect the incircle of $A B C$ at $P$ and $Q$, with $I$ between $P$ and $O$. Then $O P=r+O I$ and $O Q=r-O I$, and $\overline{P Q}$ is a diameter. Under the inversion, $P$ and $Q$ map to $P^{\prime}$ and $Q^{\prime}$ respectively. Because $P$, $I$, $O$, and $Q$ are collinear in that order, $P^{\prime}$ and $Q^{\prime}$ are diametrically opposed on $\Gamma$. It follows that the diameter of $\Gamma$ is
$$P^{\prime} Q^{\prime}=O P^{\prime}+O Q^{\prime}=\frac{R^{2}}{O P}+\frac{R^{2}}{O Q}=R^{2}\left(\frac{1}{r+O I}+\frac{1}{r-O I}\right)=\frac{2 r R^{2}}{r^{2}-O I^{2}}.$$
We plug in the values found above to arrive at $\frac{256 \sqrt{7}}{17}$.