Let $C$ be tangent to $L$ at $X$ and $O$ be the center of $C$. Let $R$ be one of points that has the problems property. Let $P,Q$ be the vertices of triangle which has the problems property. Let $X'$ be on $L$ such that $MX=MX'$.Consider the THe circle $C'$ such that be tangent to $PQR$ externally . and let $OX$ intersects $C$ at $Y$ again. we know $P$ is center of homothety which convert $C$ to $C'$ and we know point $Y$ will conwert to $X'$ in this homothety . so $P,Y,X'$ are collinear and since $Y,X'$ are constant , so the locus of $P$ is aline passes through $Y,X'$ .

Let the point of tangency of $C$ with $l$ be $T$, and the reflection of $T$ across the centre of the circle be $S$, so that $ST$ is a diameter of $C$. Let the intersection of $PS$ and $QR$ be $X$. Then by a well known lemma, $QT=RX$, and therefore $TM=MX$. Since $T,M$ are fixed, $X$ must be the reflection of $T$ through $M$, and hence the locus of $P$ is the line $SX$ such that $P$ is on the other side of $S$ from $X$. It is easy to check that this is true.

Let line L be tangent to the circle at X. Point M is fixed, then so is the reflection of X over M, say Z. The circle itself is fixed then so is the point diametrically opposite X, say Y. It is well known that P,Z,Y are collinear. And since line XY is fixed, all points P lie on this line.

[asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair P = dir(120), Q = dir(210), R = dir(330), I = incenter(P,Q,R), T = foot(I,Q,R), M = (Q+R)/2, S = Q+R-T, Tt = 2I-T; dot("$P$", P, dir(45)); dot("$Q$", Q, dir(270)); dot("$R$", R, dir(270)); dot("$T$", T, dir(270)); dot("$M$", M, dir(270)); dot("$S$", S, dir(270)); dot("$T'$", Tt, dir(240)); draw(Tt--Tt+dir(S--Tt)*1.2, red, Arrow); draw(Q--P--R, blue); draw(2Q-T--2R-S, Arrows); draw(incircle(P,Q,R)); draw(T--Tt--S, blue+dashed); [/asy][/asy] Let $T$ be the point where $C$ touches $L$, $T'$ be the point diametrically opposite $T$ on $C$, and $S$ be the reflection of $T$ over $M$. We claim that the locus is a ray with endpoint $T'$ pointing toward the direction from $S$ to $T'$, not including $T'$. It is well known that in order to satisfy the given conditions, we must have $P$, $T'$, and $S$ collinear (consider a homothety at $P$ that sends $C$ to the $P$-excircle, where $S$ is the point where the excircle touches $\overline{QR}$). Thus $P$ must lie on line $\overline{T'S}$. However, if $P$ is closer to $S$ than $T'$ or $P = T'$, it is impossible to draw the tangents $\overline{PQ}$ and $\overline{PR}$ such that $C$ is the incircle. To show that all points in the locus works, pick any $P$ and let its tangents to $C$ meet $L$ at $Q$ and $R$. Since the midpoint $M'$ of $\overline{PQ}$ is the unique point such that $T$ is the reflection of $S$ over $M'$, we must have $M'=M$, completing the proof. $\blacksquare$

Let the point $T$ be the tangency point, and $E$ be the reflection of $T$ over $M$. Then, let $D$ be the antipode of $T$. We claim that the locus is the points beyond $D$ on the ray $ED$. We first show such points work. It is well known that $PD\cap L$ is the excenter of $\triangle PQR$ (proof by homothety). Then, the midpoint of QR is the midpoint of $T$ and $PD\cap L = E$ which is $M$, so we're done. We now show that only points on this ray work. Firstly, points on the other side of $\overline{DE}$ fail since $D$ must be in the interior of $\triangle PQR$. Then, for $P\notin DE$, note that $PD\cap L= E'$ is still the excenter of $\triangle PQR$. Then, the midpoint of $QR$ is the midpoint of $TE'$ which is not equal to the midpoint of $TE$ since $E\neq E'$, so these points don't work. Thus, we 've shown that all points on the ray work, and all points not on the ray don't work, so we're done. $\blacksquare$.

Seems easy for even P1. Let $A$ be center of $C$ and $B=L\cap C$ and $B'$ be reflection of $B$ over $M$ and $B_1$ be antipode of $B$ . So, because of " Diameter of incircle Lemma" we know that $P$ should be on line $MB_1$. So, locus of $P$ is $\overrightarrow{\rm B_1X}$ where $X$ is any point on $\overrightarrow{\rm MB_1}$ such that $B_1$ lies between $M$ and $X$. The question "Why all points on this ray work?" can be easily answered like this : Take any point $P$ on this ray and draw tangents from $P$ to $C$ and intersect them with $L$ at $Q,R$. Then again because of "Diameter of Incircle Lemma" $M$ is midpoint of $QR$. So we are done!

Let $X=\mathcal{C}\cap\ell,$ $X'$ the antipode of $X$ with respect to $\mathcal{C},$ and $Y$ the reflection of $X$ over $M.$ We claim that $P$ lies on $\overline{X'Y}$ beyond $X'.$ Since $\mathcal{C}$ is the incircle of $\triangle PQR,$ we know $P$ must lie on $\overline{X'Y}$ by the Diameter of an Incircle lemma. Moreover, with this lemma, we can draw the tangents from $P$ (on $\overline{X'Y}$ beyond $X'$) to $\mathcal{C}$ to intersect $\ell$ at $Q$ and $R,$ noting that $MQ=MR.$ $\square$