Let $E,D,F$ be the tangent points of the incircle $(I)$ with the sides $CB,AB,AC,$ respectively. By the statement of the problem we cannot have $a=b$ because in this case $Q\equiv L$ for all four cases under consideration (unless we deal with a generic triangle $c=a+b$). Without loss of generality let $a>b;$ this will help us position $L$ with respect to $D$ on $AB.$ The power of $C$ wrt to the circle $(I)$ is $CP\cdot CQ=CI^2-r^2=CE^2$ and therefore $2CP^2=CE^2.$ Similarly for the power of $L$ wrt to the incircle $(I),$ which leads to $2LQ^2=LD^2.$ But $CP=LQ$ and therefore $LD=CE=s-c,$ where $s$ is the semi-perimeter. This means that the triangles $IDL,IEC$ are congruent and thus $IC=IL,$ and therefore $I$ is on the perpendicular bisector of $CL.$ a) If $CL$ is the median, then $L$ is between $D$ and $B$ and we have $\tfrac{c}{2}=AL=AD+DL=AF+CF=b$ and thus $c=2b.$ But $4CL^2=2(a^2+b^2)-c^2$ and thus $4\cdot 9CP^2=2(a^2-b^2)$ or $18(s-c)^2=2(a^2-b^2),$ or $\tfrac{9}{4}(a-b)^2=(a^2-b^2).$ For $a>b$ we get $a=\tfrac{13}{5}b.$ So this case is possible for a triangle with sides $(a,b,c)=(\tfrac{13}{5}b,b,2b)$ or its symmetric case $(a,b,c)=(a,\tfrac{13}{5}a,2a).$ b) If $CL$ is the bisector, then again $L$ is between $D$ and $B$ and thus $LD=s-c=AL-AD=\frac{bc}{b+a}-(s-a),$ which leads to $c=a+b,$ which corresponds to a generic triangle. c) If $CL$ is the altitude, then $L$ is between $A$ and $D$ and $LD=s-c=AD-AL=(s-a)-b\cdot\cos(\angle B)=s-a-b\cdot \frac{b^2+c^2-a^2}{2bc},$ which leads to $b^2=(c-a)^2,$ which again corresponds to a generic triangle. d) As above, in this case $L$ is between $D$ and $B$ and $AL=AD+LD=(s-a)+(s-c)=b$ but $AL=s-b$ and therefore $c=3b-a.$ Also, from the law of cosines in triangle $CAL$ we get $CL^2=2b^2-2b^2\cdot\frac{b^2+c^2-a^2}{2bc}.$ With $CL^2=9CP^2=\tfrac{9}{2} CE^2=\tfrac{9}{2}(s-c)^2,$ we find that $a=\tfrac{7}{3}b$ or $a=\tfrac{5}{3}b.$ The former is not acceptable because in this case $c<a-b.$ So this case is possible for a right triangle of sides $(a,b,c)=(\tfrac{5}{3}b,b,\tfrac{4}{3}b)$ or its symmetric $(a,b,c)=(a,\tfrac{5}{3}a,\tfrac{4}{3}a).$