Let $A B=z, B C=x, C A=y$. Let $X, Y, Z, O, N$ be the circumcenter of $A E F, B F D, C D E, A B C, D E F$ respectively. Note that $N$ is the nine-point center of $A B C$, and $X, Y, Z$ are the midpoints of $O A, O B, O C$ respectively, and thus $X Y Z$ is the image of homothety of $A B C$ with center $O$ and ratio $\frac{1}{2}$, so this triangle has side lengths $\frac{x}{2}, \frac{y}{2}, \frac{z}{2}$. Since $N X$ perpendicularly bisects $E F$, which is parallel to $B C$ and thus $Y Z$, we see that $N$ is the orthocenter of $X Y Z$. Moreover, $O_{1}$ lies on $Y N$ and $O_{1} X$ is perpendicular to $X Y$. To compute the area of $O_{1} O_{2} O_{3}$, it suffices to compute $\left[N O_{1} O_{2}\right]+\left[\mathrm{NO}_{2} O_{3}\right]+\left[N O_{3} O_{1}\right]$. Note that $O_{1} X$ is parallel to $N O_{2}$, and $O_{2} Y$ is parallel to $X N$, so $\left[N O_{1} O_{2}\right]=\left[N X O_{2}\right]=[N X Y]$. Similarly the other two triangles have equal area as $[N Y Z]$ and $[N Z X]$ respectively, so the desired area is simply the area of $[X Y Z]$, which is $\frac{1}{4} \frac{\sqrt{(x+y+z)(x+y-z)(x-y+z)(-x+y+z)}}{4}=\frac{\sqrt{21 \cdot 9 \cdot 5 \cdot 7}}{16}=\frac{21 \sqrt{15}}{16}$.