Define $X = AQ \cap BP$ so that we wish to show $T, X, L$ collinear. Let $N = PQ \cap CL$ be the midpoint of $\overline{CL}$ and $O$ be the midpoint of $\overline{CX}$. By Newton-Gauss, $M, N, O$ are collinear on a line $\ell$. Also, as $M, N$ are centers of $\odot(M, MC)$ and $\odot(N, NC)$, line $\ell$ is the perpendicular bisector of $\overline{CT}$. So, $OC = OT = OX \implies \angle CTX = 90^{\circ}$. But $\angle CTL = 90^{\circ}$, so $T, X, L$ are collinear as desired.

We invoke the sophisticated technique of ``haha construct parallelogram go brrr.'' Let $D$ denote the midpoint of $PQ$ and $C'$ the reflection of $C$ over $M$. Note that a homothety with radius $2$ about $C$ takes $DM$ to $LC'$, hence $LC'\parallel DM$. Now, note that because $C'$ is the reflection of $C$ about $M$, $C'$ is as far from line $DM$ as $C$, hence $L$ and $C$ are equidistant from line $DM$. By definition, $DT=DC$ and $MT=MC$, hence $DM$ is the perpendicular bisector of $CT$ and $C$ and $T$ are equidistant from line $DM$. This implies that line $LT$ is line $C'L$. Hence, it suffices to show that $C'L,AQ,BP$ concur, and we have eliminated point $T$. Now, we use barycentric coordinates to prove the result. Let $L=(x:1-x:0)$. Then, we have $P=(x:0:1-x)$ and $Q=(0:1-x:x)$. Hence, we have \[AP\cap BQ=(x^2:(1-x)^2:(1-x)x).\]Observe that $\vec{C'}=2\vec{M}-\vec{C}=(1:1:-1).$ As \[-x(1-x)\vec{C'} +\vec{L}=(x:1-x:0) + (-x(1-x):-x(1-x):(1-x)x) = (x^2:(1-x)^2:(1-x)x),\]we are done. Remark: This solution was inspired by ISL 2013 G2 and my inability to use projective techniques on this problem.

Let $CM \cap LT = K$ - then $\angle KTC = 90^{\circ}$ and $CM = MT = MK$. To prove that $AQ$, $BP$ and $KL$ concur, we shall see that $AQ$ and $BP$ divide $KL$ externally in the same ratio. Equivalently, we want $$ \frac{S_{AKQ}}{S_{ALQ}} = \frac{S_{BKP}}{S_{BLP}}. $$On the other hand, from the parallelogram $PCQL$ we have $S_{ALQ} = S_{CLQ} = S_{CLP} = S_{BLP}$. Analogously the parallelogram $ACBK$ gives $S_{AKQ} = S_{BKP}$ and so we are done.

Let $N=CL\cap PQ$ and $X=AQ\cap BP$. Claim: $MN\parallel LX$ Proof: Moving points! Fix parallelogram $PCQL$ and animate $A\in CP$ with degree 1. Then by Zack's lemma: $$\deg(B)=\deg(AL\cap QC)=1$$$$\deg(X)=\deg(AQ\cap PB)\leq \deg(A)+\deg(Q)+\deg(P)+\deg(B)=2$$$$\deg(M)=\deg(\frac{A+B}{2})=2$$Since $\deg(MN), \deg(LX)\leq 2$, it suffices to verify the claim for two choices of $A$. Take $A=P$ and $A=\infty_{AC}$. $\blacksquare$ Now let $T$ be the intersection of the circle centered at $M$ with radius $CM$ and the circle centered at $N$ with radius $CN$. Clearly $LT \perp CL$, but also $CT\perp MN\parallel LX$. Hence $L-X-T$ are colinear, as desired.

Rather neat result. We denote by $R$ the intersection of lines $AQ$ and $BP$ and by $N$ the midpoint of $LC$. First, note that by the Newton-Gauss Theorem on self-intersecting quadrilateral $AQBP$, the midpoints of segments $AB$ , $PQ$ and $CR$ must be collinear. Thus, by the Midpoint Theorem it follows that $LR \parallel MN$. Now, it is well known that the radical axis is perpendicular to the line joining the centers of two circles. Thus, $MN \perp CT$. Further, $LT \perp CT$ as well so $LT \parallel MN$. But this means that $\overline{LT}$ passes through point $R$ as desired.