A very nice and hard problem Here's a quick sketch: Let $X=KL \cap \ell$ and $T$ be the center of $(IPQ)$. The key claim is that the $XT$ passes through the midpoint of major arc $\widehat{BC}$ in $(ABC)$ (let it be $N$), then we can do extraversion. 1. Prove that $AI$ is tangent to $(IPQ)$ by angle chasing. 2. Prove that $XN$ bisects $AK$. Let $W=XN\cap AK$. By noticing that $$-1=N(AK;XL)=(AK;W\infty_{AK})$$ since $AK \parallel NL$ we prove this claim. 3. Let $H$ be the foot of perpendicular from $A$ to $BC$ which lies on $\ell$ by definition. Obviously $W$ is the center of $(AHK)$. Now we can get rid of $X$ and prove that $WTN$ are collinear. The final claim is $(IPQ)$, $(AHK)$ and $(N,NB)$ are coaxial which finishes the problem. By using coaxial lemma we need to show $$\frac{AI^2}{-AB \cdot AC}=\frac{HP \cdot HQ}{-HB \cdot HC}=\frac{KI^2}{-KB \cdot KC}$$ But they are all $\tan B/2 \tan C/2$ and we are done. $\blacksquare$

Let $\ell$ cut $UV$ at $W$, denote $U,V$ as the center of $(IPQ), (JRS)$ respectively then by simple angle chasing we get $UQ \parallel VS, UP \parallel VR$ therefore $\dfrac{WQ}{WS} = \dfrac{UQ}{VS} = \dfrac{UP}{VR} = \dfrac{WP}{WR}$ therefore $WP \cdot WS = WQ \cdot WR$ therefore $W$ lies on the radical axis of $(KQR)$ and $(KPS)$. We will prove $KL$ is indeed the radical axis of $(KQR)$, $(KPS)$ Let $LK$ cut $(O)$ at $G \neq L$, we will prove $G$ lies on $(KQR)$ and similiarly with $(KPS)$. Let $AI$ cut $(O)$ at $X \neq A$, $XL$ cut $BC$ at $Y$. Consider an inversion about a circle at $K$ with radius $\sqrt{KB\cdot KC}$, it sends $G \mapsto L$, $\ell \mapsto (KXY)$, $Q \mapsto Q', R \mapsto R'$ then $Q', R'$ is the intersection of $(KBJ), (KIC)$ with $(KXY)$ respectively. We will need to prove $R',Q',L$ are collinear. We have $XI^2 = XB^2 = XL \cdot XY$, therefore $\dfrac{LX}{LY} = \dfrac{XL \cdot XY}{YL \cdot XY} = \dfrac{XI^2}{YB \cdot YC}$$(*)$ By simple angle chasing we get $\triangle{R'IX} \sim \triangle{R'CY}$ and $\triangle{Q'JX} \sim \triangle{Q'BY}$ therefore $\angle XR'Y = \angle IR'C = \angle JKC = 180^\circ - \angle XQ'Y$ therefore $XYR‘Q'$ are cyclic. Therefore let $R'Q'$ cut $XY$ at $L'$ then $\dfrac{L'X}{L'Y} = \dfrac{R'X}{R'Y} \cdot \dfrac{Q'X}{Q'Y} = \dfrac{XI}{CY} \cdot \dfrac{XJ}{YB} = \dfrac{XI^2}{YC \cdot YB}$. Combine with $(*)$ we get $\dfrac{L'X}{L‘Y} = \dfrac{LX}{LY}$ therefore $L \equiv L'$ therefore $L$ lies on $R'Q'$. Hence the problem is done.

WLOG $\angle B > \angle C$, $O_1$ is the circumcenter of $(IPQ)$, $I'$ is the antipode of $I$ in $(IPQ)$ and $E = IO_1 \cap (PO_1Q)$. $O_2$ is the circumcenter of $(JRS)$, $J'$ is the antipode of $J$ in $(JRS)$ and $F = JO_2 \cap (RO_2S)$. $D$ is the foot of $A$ onto $BC$, $G$ is the foot of $J$ onto $BC$. Main Claim: $E$ and $F$ are on $KL$. Proof. $ABC \cup I \sim FRS \cup J \implies \frac{JF}{AI} = \frac{DG}{r}(*)$ and $\frac{DG}{ML} = \frac{DG}{DK} \cdot \frac{DK}{ML} = \frac{AJ}{AK} \cdot \frac{AK}{AL} = \frac{AJ}{AL} = \frac{AD}{AI} (**)$. If $r$ is the radius of incircle, $$F \in KL \iff \frac{KM}{ML} = \frac{KJ}{JF} \overset{*} \iff \frac{DG}{ML} = \frac{KJ}{KM} \cdot \frac{r}{AI} \overset{**}\iff \frac{AD}{AI} = \frac{r}{AI} \cdot \frac{KA}{KI} \iff \frac{AD}{r} = \frac{KA}{KI}$$ which is true. Similar approach gives $E \in KL$. $\square$ There exists a homothety taking $IQP \cup O_1 \cup E$ to $J'SR \cup O_2 \cup F$ which gives $O_1O_2$, $l$ and $KL$ are concurrent. $\blacksquare$

Let $LK$ meet $l$ at $X$. We make the stronger claim that $X$ is the internal homothetic center of the two circles. Let $II'$ and $JJ'$ be diameters in $(IPQ)$ and $(JRS)$. Lemma: Let triangles $AP_1P_1'$, $AP_2P_2'$, and $AP_3P_3'$ all share angle bisector $AD$. Then there exists a conic passing through $P_1$, $P_2$, $P_3$, $P_1'$, $P_2'$, and $P_3'$.

Notice that $AK$ is the bisector in $ABC$, $ALX$, and $AIJ$ * This last fact is not immediate and follows from $IK/JK=IA/JA$. thus by the lemma there exists a conic $\mathcal{C}$ passing through $B$, $C$, $I$, $J$, $L$, and $X$. Let $\mathcal{C'}$ be the circumconic that is the isogonal conjugate of line $IJ$ in triangle $BCL$. As $I$ and $J$ are an isogonal pair in this triangle, $\mathcal{C'}$ must pass through these two points. As a conic is uniquely determined by $5$ points we must have that $\mathcal{C}=\mathcal{C'}$. Notice that the isogonal conjugate of $A$ in this triangle is the point at infinity perpendicular to $BC$ thus one of the axes of $\mathcal{C}$ must be perpendicular to $BC$. As $l$ is also perpendicular to $BC$ we must have that $X$ is the unique intersection point of $\mathcal{C}$ with line $l$. Applying Desargues’ Involution Theorem on quadrilateral $IBJC$, conic $\mathcal{C}$, and line $l$ gives that their is an inversion centered at $X$ swapping $(P,S)$ and $(Q,R)$ or that $XP/XR=XQ/XS$. Notice quadrilaterals $IPI'Q$ and $J'RJS$ have parallel sides and diagonals so must have an internal homthetic center. As $P$, $Q$, $R$, and $S$ are collinear, this center must lie on line $l$ and must be $X$ by the previously mentioned ratios.

See a generalization of this problem here (Problem 9).

Solution 1. Let $O_1, O_2$ be the centers of circles $(IPQ)$ and $(JRS)$ respectively. Let $O_1O_2$ intersect $\ell$ and $AI$ at $F$ and $G$ respectively. We will prove the problem with a few steps: Step 1: $G$ is the exsimilicenter of the two circles $(IPQ)$ and $(JRS)$. Proof: Note that by angle chasing, $\angle IPQ=\angle PAB+\angle PBA=90^{\circ}-\angle ABC +\frac{1}{2}\angle ABC=180^{\circ}-\angle AIC=\angle AIQ$. This implies $AI$ is tangent to $(IPQ)$. In a similar fashion, we can prove that $AJ$ is tangent to $(JRS)$ as well. Hence $AI$ is a common external tangent of the circles, and hence $G$ must be their exsimilicenter. $\square$ Step 2: $F$ is the insimilicenter of the two circles $(IPQ)$ and $(JRS)$. Proof: This follows from the equality $\angle PIQ+\angle RJS=180^{\circ}$. To see this more clearly, let $IX$ and $JY$ be diameters in $(IPQ)$ and $(JRS)$ respectively. As $IJ$ is their common tangent, the negative homothety sending $(IPQ)$ to $(JRS)$ must also send $I$ to $Y$ and $X$ to $J$. Till this end, let us observe that $IPXQ$ and $YSJR$ are homothetic quadrilaterals in this orientation, as they have parallel sides and parallel diagonals. Hence the center of negative homothety lie on the line $PQ\equiv RS$, and line $O_1O_2$ as well. This point must then be $G$. $\square$ Step 3: $G$ is the midpoint of $AK$. Proof: Let $\ell$ intersect $BC$ at $H$. The claim is that the circle $(HAK)$ is coaxial to both circles $(IPQ)$ and $(JRS)$, as this will prove that the center of the three circles are colinear, namely, the points $G, O_1, O_2$ are colinear. This can be proved by invoking the Coaxial Circles Lemma: If we write $\text{Pow}(T, \omega)$ as the power of a point $T$ with respect to $\omega$, then it suffice to prove that $$\frac{\text{Pow}(H, (IPQ))}{\text{Pow}(H, (JRS))}=\frac{\text{Pow}(A, (IPQ))}{\text{Pow}(A, (JRS))}=\frac{\text{Pow}(K, (IPQ))}{\text{Pow}(K, (JRS))}$$ The first equality is more complicated: Observe that $(A,H;P,R)=(A,K;I,J)=-1=(A,H;Q,S)$ by projecting from $B$ and $C$ respectively. This implies $$\frac{\text{Pow}(H, (IPQ))}{\text{Pow}(H, (JRS))}=\frac{HP}{HR}\cdot\frac{HQ}{HS}=\frac{AP}{AR}\cdot\frac{AQ}{AS}=\frac{\text{Pow}(A, (IPQ))}{\text{Pow}(A, (JRS))}$$ The second equality follows from $(A,K;I,J)=-1$, because $$\frac{\text{Pow}(A, (IPQ))}{\text{Pow}(A, (JRS))}=\left(\frac{AI}{AJ}\right)^2=\left(\frac{KI}{KJ}\right)^2=\frac{\text{Pow}(K, (IPQ))}{\text{Pow}(K, (JRS))}$$ which proves the lemma. $\square$ With these preparations, let's finally prove that $F, K, L$ are colinear. Let $Z$ be the projection of $F$ to $AI$, then we get $-1=(G,F;O_1,O_2)=(G,Z;I,J)$ by projecting along the parallel lines in the direction perpendicular to $AI$. This implies that $Z$ is the image of $G$ under inversion with respect to the circle with diameter $IJ$. Moreover, we also observe that the circle with diameters $IJ$ and $AK$ must be orthogonal as $(A,K;I,J)=-1$, so $Z$ lies on their common radical axis. This implies that if $N$ is the midpoint of $IJ$, that is the midpoint of minor arc $BC$, then $Z$ must also be the image of $N$ under inversion with respect to the circle with diameter $AK$, and hence $(A,K;Z,N)=-1$. Finally, as the triangles $\triangle AFZ$ and $\triangle ALN$ are similar, we get $$\frac{ZK}{ZN}=\frac{AK}{AN}=\frac{ZF}{LN}$$ which implies $F, K, L$ are colinear, as desired. $\blacksquare$ Comment 1: In fact, the circles $(IPQ)$ and $(JRS)$ are inverses of each other with respect to circle $(HAK)$.

Solution 2: This time we will show that $O_1O_2$ passes through $M$, the midpoint of major arc $BAC$. Then we will invoke Step $3$ in the Solution $1$ to finish. Let $U$ be the second intersection of $(IPQ)$ with $(BICJ)$, and let $V$ be the second intersection of $(JRS)$ with $(BICJ)$. Claim: The lines $IU, JV, BC$ are concurrent. Proof: Define the point $H$ as in Solution $1$. Note that $U$ is the Miquel point of the complete quadrilateral $BHCIPQ$, as $U$ is the intersection of the circles $(BIC)$ and $(IPQ)$. This immediately implies $BUPH$ is cyclic. We then angle chase that: $$\angle(HU,UI)=\angle(HU, UP)+\angle(PU,UI)=\angle(HB,BP)+\angle(PI,IK)=\angle(HK,KI)$$ which gives $UIKH$ is cyclic. In a similar fashion we can prove that $VJKH$ is cyclic, hence by Radical Axis Theorem on $(UIKH), (VJKH)$ and $(UIJV)$ implies the claim. $\square$ With the claim, we can prove that $O_1, O_2, M$ are colinear: We simply note that because $O_1I$ and hence $O_1U$ are tangents to $(BICJ)$, then $IU$ is the polar of $O_1$. Likewise $VJ$ is the polar of $O_2$, while $BC$ is the polar of $M$ in this circle. Hence by the claim implies $O_1, O_2, M$ must be colinear. Till this end, let's invoke Step $3$ (without Step $2$) of Solution $1$. Suppose we redefine $F$ to be $KL$ intersect $AH$ instead, and let $G$ be the intersection of $FM$ and $AK$, then we want to prove that $G$ is the midpoint of $AK$, so that $G$ is the same point as defined in Solution $1$ and line $O_1O_2$ is exactly the line $GM$. Let $AM$ intersect $KL$ at $W$, then as the lines $AF$ and $AL$ are isogonal, then $AK$ and $AW$ are the internal and external angle bisector of $\angle FAL$. This implies $(F,L;K,W)=-1$. Moreover, $AK$ is parallel to $ML$ because if $N$ is the midpoint of minor arc $BC$ then $ANLM$ is a rectangle. Hence by projecting from $M$ we get $(A,K;G,\infty_{AK})=-1$, which implies $G$ must be the midpoint of $AK$ as desired. This completes the proof. $\blacksquare$

Solution 3: This solution will not require us to prove $G$ is the midpoint of $AK$, but rather just the Claim in Solution $2$, as we will see below. Following the same constructions for $U$ and $V$ as in Solution $2$, the claim implies that $IU$, $JV$ and $BC$ are concurrent at a point $D$. Suppose we let $N$ be the midpoint of minor arc $BC$, and the lines $DA$, $DN$ intersect the circumcircle $(ABC)$ again at $E$ and $T$ respectively. Claim 1: $T$ lies on $O_1O_2$. Proof: We claim that $O_1INTU$, $O_2JNTV$ and $NKHT$ are cyclic, but this is simply because $UIHK$ is cyclic (as shown in Solution $2$ via angle chasing), so we get $DT\cdot DN=DB\cdot DC=DV\cdot DJ=DU\cdot DI=DH\cdot DK$. With this, we can apply Reim's Theorem on the cyclic quadrilaterals $O_1INT$ and $O_2JNT$ to see that $O_1$, $O_2$ and $T$ are colinear. $\square$ Now we are ready to solve the problem. First, observe that $AEHK$ is cyclic because $DE\cdot DA=DB\cdot DC=DH\cdot DK$, and so $\angle AEK=\angle AHK=90^{\circ}$, that is the points $E$, $K$ and $L$ are colinear. Let us also construct $G$ to be the intersection of $O_1O_2$ with $AI$. We will need just one more claim. Claim 2: The quadrilaterals $AEHK$, $ETKG$ and $AGHT$ are cyclic. Proof: The key observation (that bypasses the need to prove $G$ being midpoint of $AK$) is that $\angle GTN=\angle O_1IN=90^{\circ}$. This characterizes $G$ uniquely without reference to $O_1$ and $O_2$. - $ETGK$ is cyclic: $\angle TEK=180^{\circ}-\angle TNL=90^{\circ}-\angle TNG=\angle TGK$ - $AGHT$ is cyclic: $\angle GTH=90^{\circ}-\angle HTN=90^{\circ}-\angle AKH=\angle HAG$ - $AEHK$ is cyclic has been proved before. This proves Claim 2. $\square$ By Claim $2$, the lines $GT$, $EK$ and $AH$ are concurrent by applying Radical Axis Theorem on the three circles $(AEHK)$, $(ETKG)$ and $(AGHT)$. This concludes the proof. $\blacksquare$ Comment 2: The point $T$ actually coincides with the famous point $X$ in IMO Shortlist 2011 G4.

Solution 4: (Wong Jer Ren) It is possible to solve the problem without requiring to prove $O_1O_2$ passes through any other point. Let $LK$ intersect $AH$ at $F$ and $F'$ be the insimilitude center of $(O_1)$, $(O_2)$. Let line $LK$ intersect $(ABC)$ again at $E$. Claim : $QKRE$ cyclic Proof: Denote $R'$, $Q'$, $E'$ as the images of $R$, $Q$, $E$ under the $\sqrt{bc}$-inversion at $A$ wrt triangle $ABC$. It suffices to show $R'E'Q'N$ cyclic. First, we note some properties that follow immediately by the inversion: $\bullet$ $A$, $R'$, $L$, $Q'$ collinear $\bullet$ $E'$ is the intersection of $LN$ intersect $BC$ $\bullet$ $AIR'C$ cyclic $\bullet$ $ABJQ'$ cyclic Subclaim 1: $LR'=LC$ Proof: We have $\angle LR'C= 180^{\circ}-\angle AR'C' = 180^{\circ}-\angle AIC = 90^{\circ} - \frac{1}{2}\angle ABC$, while $\angle R'LC=\angle ALC=\angle ABC$. $\square$ Similarly, $LQ'=LB$. Hence, it suffices to show $LB \cdot LC=LN \cdot LE'$. Subclaim 2: $\triangle LBE'\sim \triangle LNC$ Proof: We have $\angle LBE'=\angle LBC=\angle LNC$, while $\angle BLE'=180^{\circ}-\angle BLN=180^{\circ}-\angle BAN=180^{\circ}-\angle CAN=\angle NLC$. Therefore $\displaystyle\frac{LB}{LE'}=\frac{LN}{LC}$, which implies $LB \cdot LC=LN \cdot LE'$ and we're done. $\square$ This finishes the proof of claim. $\square$ Similarly, $PKSE$ cyclic. Therefore $FP \cdot FS=FK \cdot FE=FR \cdot FQ$, which implies $$\frac{FP}{FQ}=\frac{FR}{FS}$$ However, $F'$ lies on $AH$ too from Step $2$ of Solution $1$ and $$\frac{F'P}{F'Q}=\frac{F'R}{F'S}$$ Hence, $F=F'$, which implies $F$ lies on $O_1O_2$. $\blacksquare$ Remark: It can be proven that if $\ell$ is any arbitrary line through $A$, then the problem remains true if we replace $L$ to be a point on $(ABC)$ such that $\ell$ and $AL$ are isogonal w.r.t $\angle BAC$.