Let $ A_3 $ $ \equiv $ $ \odot (O) $ $ \cap $ $ \odot (AA_1A_2). $ From Reim's theorem we get $ A_2A_3 $ passes through the midpoint $ A_4 $ of arc $ BC $ in $ \odot (O), $ so the intersection $ T $ of $ IO, $ $ A_2A_3 $ is the exsimilicenter of $ \odot (I) $ $ \sim $ $ \odot (O), $ hence notice $ \tfrac{TA_2}{TA_4} $ $ = $ $ \tfrac{r}{R} $ we conclude that $$ \text{Power}(T,\odot(AA_1A_2)) = \frac{r}{R} \cdot \text{Power}(T,\odot(O)). $$

Essentially the same solution, but I will write anyways.

[asy][asy] size(8cm); pointpen=black; pathpen=heavygrey; pointfontpen=fontsize(10); pair A=dir(130), B=dir(215), C=dir(-35), I=incenter(A,B,C), A_1=extension(A,I,B,C), A_2=IP(D(incircle(A,B,C),red),B--C), M=(0,-1), A_3=IP(L(M,A_2,-0.1,5),D(unitcircle,heavygreen)), J=I/(1-inradius(A,B,C)), O=(0,0); D(circumcircle(A,A_1,A_2)); D(arc(M,abs(M-I),degrees(B-M)+10,degrees(C-M)-10,CW),dashed+blue); DPA(A--B--C--cycle^^O--J); D(M--A_3,dashed+blue); D("A",A,A); D("B",B,B); D("C",C,C); D("J",J,J); D("I",I,NW); D("M",M); D("O",O,NW); D("A_1",A_1); D("A_2",A_2); D("A_3",A_3,NW); [/asy][/asy] Another perspective is to consider an inversion in $\odot(M,MI)$, where $M$ is the midpoint of arc $\widehat{BC}$, which fixes $(AA_1A_2)$, so that if $A_3=(AA_1A_2)\cap(ABC)$, then $\overline{MA_2A_3}$ passes through the exsimilicenter $J$. Then since $\frac{JA_1}{JM}=\frac{r}{R}$, $$\mathrm{pow}(J,(AA_1A_2))=JA_2\cdot JA_3=\frac{r}{R}\cdot\mathrm{pow}(J,(ABC)).$$

Take the point $X_{56}$, the exsimilicenter of circumcircle and incircle. Let $M_a$ be the midpoint of arc $BC$, if let $M_aA_1\cap (ABC)$ again at $A_3$ then by the Shooting Lemma, $A_3\in (AA_1A_2)$. Also observe $X_{56}\in M_aA_1$, and $pow(X_{56},(AA_1A_2))=X_{56}A_1\cdot X_{56}A_3=X_{56}M_a\cdot X_{56}A_3\cdot \frac{r}{R}=pow(X_{56},(O))\cdot \frac{r}{R}$ which is fixed. By symmetry, this is the power of $X_{56}$ wrt $(BB_1B_2)$ and $(CC_1C_2)$ as well, thus its radical center. Clearly $X_{56}\in OI$, so we are done.

Construct a circle $X$ with diameter $AI$. By radical axis and the fact that the mixtilinear touch point with $(O)$ is on $(M_aA_1A_2)$, we have that $A_3$ is the intersection of $X$ and $(O)$. since the spiral similarity at $A_3$ sending $B_2C_2$ to $CB$ sends $I$ to $M_a$ and the foot of the $A_2$ altitude onto $B_2C_2$ and $A_3, I$ are collinear with the mentioned foot of altitude, $M_a, A_2, A_3$ are collinear. Also if the A mixtilinear incircle touches the circumcircle at $T$, then we claim that the radical center is precisely the intersection of $AT, M_aA_3$. It is easily seen that the mixtilinear cevians intersect on $OI$ by the three homotheties theorem, at the exsimilicenter of the incircle and the circumcircle. This implies our claim.

Take $T$ as the center of homothety from incircle to excircle.It is easy to see the power of this point is same W. r. t the three circles. Since it lies on $OI$ hence done !!

Here is my solution for this problem Solution

Let $J$ be the external homothetic center of ($I$) and ($O$); $AJ$ $\cap$ ($AA_1A_2$) = {$A$, $D$}, $AJ$ $\cap$ ($O$) = {$A$, $X$}, $BJ$ $\cap$ ($BB_1B_2$) = {$B$, $E$}, $BJ$ $\cap$ ($O$) = {$B$, $Y$}; $AA_1$ $\cap$ ($O$) = {$A$, $M$} It's easy to see that: $J$ $\in$ $IO$ We have: $\widehat{ADA_2}$ = $180^o$ $-$ $\widehat{AA_1A_2}$ = $180^o$ $-$ $\dfrac{1}{2}$ ($\stackrel\frown{AB}$ + $\stackrel\frown{MC}$) = $180^o$ $-$ $\dfrac{1}{2}$ $\stackrel\frown{ABM}$ = $\dfrac{1}{2}$ $\stackrel\frown{ACM}$ = $\widehat{AXM}$ Hence: $DA_2$ $\parallel$ $XM$ or $D$ $\in$ ($I$) Similarly: $E$ $\in$ ($I$) So: $\overline{JD}$ . $\overline{JA}$ = $k$ $\overline{JX}$ . $\overline{JA}$ = $k$ $\overline{JY}$ . $\overline{JB}$ = $\overline{JE}$ . $\overline{JB}$ ($k$ = $\dfrac{\overline{JI}}{\overline{JO}}$ ) It leads to: $P_{J / (AA_1A_2)}$ = $P_{J / (BB_1B_2)}$ Similarly: $P_{J / (BB_1B_2)}$ = $P_{J / (CC_1C_2)}$ Therefore: $P_{J / (AA_1A_2)}$ = $P_{J / (BB_1B_2)}$ = $P_{J / (CC_1C_2)}$ or $J$ is radical center of these 3 circles.