Assume WLOG that the line $ l$ cuts the sides $ CA$, $ AB$ in their interior. According to Miquel's theorem, applied to the complete quadrilateral $ BCEF$, the circumcircles of triangles $ AEF$, $ BFD$, $ CDE$ are concurrent on the circumcircle of $ ABC$. Let this point be $ M$. On other hand, since $ l$ passes, for example, through the intersection of the circles $ AEF$, and $ BFD$, its reflection in the line $ O_{1}O_{2}$ will pass through $ M$. Thus, we conclude that the reflections of $ l$ in the sidelines of triangle $ O_{1}O_{2}O_{3}$ are concurrent. The conclusion now follows from the converse of the "Anti-Steiner" theorem.

Here's another approach: WLOG the points $ D, F, E$ lie in that order in the line $ l$ It is well known that circumcircles of $ AFE, BFD, CDE$ meet are concurrent at say, point $ M$. Then $ O_{1}$, $ O_{2}$ and $ O_{3}$ are circumcentres of triangles $ MFE, MFD, MDE$, and we want to prove orthocentre of $ O_{1}O_{2}O_{3}$, $ H$, lies on line $ DEF$ - so now we can ignore the points $ A, B, C$. It's easy to see that $ O_{1}O_{3}$ is perpendicular to $ ME$ (since $ O_{1}M = O_{1}E, O_{3}M = O_{3}E$), so $ O_{2}H$ is perpendicular to $ O_{1}O_{3}$, so $ O_{2}H$ is parallel to $ ME$ and $ O_{1}H$ is parallel to $ MD$. Let $ H'$ be the point where the circumcircle of $ O_{1}O_{2}F$ meets line $ DFE$ again - it suffices to prove that $ H'$ and $ H$ coincide, i.e. $ O_{2}H', ME$ and $ O_{1}H', MD$ are both pairs of parallel lines. But this is a simple angle chase using the cyclic quad $ O_{2}FO_{1}H'$ : we have $ \angle O_{2}H'F = \angle O_{2}O_{1}F = \frac {\angle MO_{1}F }{2} = \angle MEF$, so $ O_{2}H$ and $ ME$ are parallel as required; similarly $ O_{1}H'$ and $ MD$ are parallel.

Call $ H$ is the orthorcenter of $ \Delta O_{1}O_{2}O_{3}$ And the circumcircle of $ \Delta ABC$ is $ (O)$ We know that four circle: $ (O), (O_{1}), (O_{2}), (O_{3})$ have one common point, call it is $ T$ Easy to proof that: $ (DO_{2},DO_{3})\equiv(O_{1}O_{3},O_{1}O_{2})(mod\pi)$ $ \equiv(HO_{2},HO_{3})(mod \pi)$ $ \Rightarrow H, O_{2}, O_{3}, D$ lie on a circle Similar, we have $ H, O_{2}, O_{1}, F$lie on a circle $ \Rightarrow$ $ (HF, HD)$ $ \equiv(HF, HO_{1}) + (HO_{1},HO_{3}) + (HO_{3}, HD)(mod\pi)$ $ \equiv(O_{2}F, O_{2}O_{1}) + (O_{2}O_{3},O_{2}O_{1}) + (O_{2}O_{3}, O_{2}D)(mod\pi)$ $ \equiv 0 (mod\pi)$ We get the result

My solution: $\odot (\triangle AEF),\odot (\triangle BFD),\odot (\triangle CED)$ are concurrent at $M$ miquel point of complete quadrilateral $EFBC$ it's well known that $MO_1O_2,O_3$ is cyclic quadrilateral. note that $O_1O_2,O_2O_3$ are perpendicular bisectors of $MF,MD$ let $O_1O_2\cap MF=N,O_2O_3\cap MD=S$ so $NS$ is Simson line of $\triangle O_1O_2O_3$ and also it is midline of $\triangle MFD$ so if $H$ is orthocenter of $\triangle O_1O_2O_3$ line $PH$ must bisect $SN$ so $H$ lies on $FD$. DONE

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