Actually they are concurrent at $OI$

Hint

Let $I_b,I_c$ be the excenters of $\triangle ABC$ againts $B,C.$ Incenter $I$ and circumcicle $(O)$ of $\triangle ABC$ become orthocenter and 9 point circle of $\triangle I_aI_bI_c$ $\Longrightarrow$ $(O)$ cuts $\overline{I_bI_c}$ again at its midpoint $U.$ $I_aD$ is the polar of $I$ WRT the circumcircle $(U)$ of $BCI_bI_c$ $\Longrightarrow$ $IU \perp DI_a,$ i.e. $A'$ lies on the line connecting $I$ with the midpoint $U$ of the arc $BAC$ of $(O)$ $\Longrightarrow$ $A'$ is the tangency point of the A-mixtilinear incircle $\omega_A$ of $\triangle ABC$ with $(O).$ For a proof, see problem 2.20 of 2005 mosp 2.20(g) 4.44(g). Thus, $A$ and $A'$ are the exsimilicenters of $\omega_A \sim (I)$ and $\omega_A \sim (O)$ $\Longrightarrow$ $AA'$ passes through the exsimilicenter $X_{56}$ of $(I) \sim (O).$ Likewise, $BB'$ and $CC'$ pass through $X_{56}.$

Dear Mathlinkers, If I am not wrong , the point of concurs is the Schroder point. We can see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=283185 Sincerely Jean-Louis

if $L$ is the midpoint of arc $BAC$ then $AI.AI_a=AB.AC=AL.AD$ (similarity) and so we have $AIL \sim ADI_a$ and so $LI \perp DI_a$ but it is well-known that $LI$ passes thru the tangent point of $A$-mixtillinear circle with $(ABC)$.the rest is trivial.

The concurrency point is also the isogonal conjugate of $\text{Nagel Point}$ of $\triangle ABC$.

My solution : Since $ I $ is the intersection of $ BI_b $ and $ CI_c $ , so $ I_aD $ is the polar of $ I $ with respect to $ \odot (BCI_bI_c) $ , hence the line through $ I $ and perpendicular to $ I_aD $ passes through the midpoint of $ I_bI_c $ . i.e. $ A' $ is the tangency point of the A-mixtillinear circle with $ \odot (ABC) $ Similarly, $ B', $ $ C' $ is the tangency point of B- mixtillinear circle, C- mixtillinear circle with $ \odot (ABC) $ , so we conclude that $ AA', $ $ BB', $ $ CC' $ are concurrent at the exsimilicenter of $ \odot (I) $ $ \sim $ $ \odot (O) $ ($ X_{56} $ in ETC) . Q.E.D

Claim 1 Let $D$ be the midpoint of arc $CAB$ and $I$ the incenter of the triangle. Than $D$,$I$ and the point of touch of the $A$-mixtlinear circle are collinear Proof: Let the mixtlinear circle touch $AB$,$AC$ in $P,Q$ and the circumcircle in $S$. It s well know that $I$ is the midpoint of $PQ$ The homothety taking the $A$-mixtlinear circle to circumcircle (at $S$) takes $P$,$Q$ to the midpoints of arcs $AB,AC$,(X,Y respectively) and clearly it takes $L$ to $A$. Now by inspection $SL$ is the symmedian in $\triangle PQS$ $\implies$ $AS$ is the symmedian in $\triangle SXY$.Now let $SI$ sut the circumcircle in $D^{'}$.We will prove $D\equiv D^{'}$ From the properties of the median and symmedian $AP=D^{'}Q$ but $AP=DQ$ so we are done. Claim 2: Let the respective mixtlinear touch the $\odot ABC$ in $A^{'},B^{'},C^{'}$ than $AA^{'}$,$BB^{'}$,$CC^{'}$are concurrent Proof $A^{'}$ is the center of positive homothety that takes $A$-mixtliniear circle to $\odot ABC$ $A$ is the center of positive homothety that takes the aforementioned circle to the inscribed circle thus by Monge -De' Alambert $AA^{'}$ passes thru the positive center of homothety that takes the incircle to the $\odot ABC$ thus we are done Now let $I_{A},I_{B},I_{C}$ be the respective excenters. Let the midpoint of arc$CAB$ be $F$ Its well known that $\triangle ABC$ is the orthic triangle of $\triangle I_{A}I_{B}I_{C}$ By La -Hire s theorem $I_{A}D$ is the polar of $I$ wrt $\odot I_{C}I_{B}BC$ so $FI$($F$ is the midpoint of $I_{B}I_{C}$) is perpendicular to $I_{A}D$ and so we are done by Claim1 and Claim2

Let the incircle touch $BC$ at $X$ and the projection from $I$ onto $DI_a$ be $F_a$. Thales' yields that $AIXF_aD$ and $BICI_aF_a$ are cyclic. Now, a well-known lemma implies $F_a$ is the reflection of $I$ over the $A$-Mixtilinear touch point, i.e. the $A$-Mixtilinear incircle touches $(ABC)$ at $A'$. Now, symmetry implies $AA', BB', CC'$ concur at $X_{56}$. $\blacksquare$ Remark: As I was working on another question from this exam, I read this problem's statement and realized that it's strongly related to ISL 2020/G6. Also, the well-known lemma I cited is linked here!

Let $L$ be the midpoint of arc $BAC$ and $LI \cap (BIC)=I,K$. $LK \perp KI_a$ so by radical axis theorem on $(ABC),(LI_a),(BIC)$ we have that $KI_a \cap BC=D$. This implies $A',B',C'$ are the $A,B,C$ mixtilinear touch points so the lines concur at isogonal conjugate of the nagel point.