A BEAUTEOUS problem! Suppose the center of $\Omega$ is $O$ and center of $\omega$ is $I$. The insimilicenter of $\Omega$ and $\omega$ is $S$ which is the intersection of $OI$ and $AB$. $N$ is the perpendicular foot of $S$ on $AI$. $S$ is a point on the plane such that $\triangle AOM\stackrel{+}{\sim}\triangle AIL$. $SK\cap AC=T$. It is obvious that $\triangle ASN\stackrel{+}{\sim}\triangle AIL$. Then $\triangle AIS\stackrel{+}{\sim}\triangle ALN$. Let $IS\cap NL=U$. Because $$\measuredangle AIU=\measuredangle NIS=\measuredangle TLN=\measuredangle ALU,$$we have $A,\,I,\,L,\,U$ are concyclic. Analogously $A,\,S,\,N,\,U$ are concyclic. Then $$\measuredangle IUL=\measuredangle IAL=\measuredangle SAN=\measuredangle SUN$$which tells us that $U,\,L,\,N$ are collinear. Analogously we can know that $U,\,L,\,N,\,M$ are collinear. As $\measuredangle INS=\measuredangle IKS=\measuredangle ILA=90^\circ$, we have $I,\,K,\,N,\,S$, and $I,\,K,\,T,\,L$ are concyclic. Hence $$\measuredangle ITK=\measuredangle ILK=\measuredangle IAU=\measuredangle NSU=\measuredangle LKI$$combine with $\measuredangle IKT=90^\circ$ we know that $\measuredangle TIK=\measuredangle LKT$. Thence $$\measuredangle SKN=\measuredangle SIN=\measuredangle KLT=\measuredangle KIT=\measuredangle TKL$$stands for $L,\,K,\,N$ are collinear. So $L,\,K,\,M$ are collinear. As $A,\,O$ and $|\angle BAC|$ are invariant, $M$ is a fixed point. And then we've done.

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