3-point' curveIf you have a loop of string, a fixed point and a pencil, and...

Step 1
Let length of sting be $>max(AB,BC,CA)$. If the given three points are A,B and C are not in a straight line, then
there are three possibilities to trace three ellipses and the locus is composed of the these three discontinuous ellipses.
Step 2
Center point B can be ignored when inter-focal distance is AC,
Center point C can be ignored when inter-focal distance is BA, and
Center point A can be ignored when inter-focal distance is BC.
So in effect each is only a 2-point curve.

Step 1
Assume that three non-collinear points A, B, C in the plane are given, and that you have a loop of string of length $\ell >|AB|+|BC|+|CA|$. Slinging this string around the three points and a pencil you can draw a loop $\gamma$ around $\mathrm{△}(ABC)$ in the obvious way, keeping the string tight at all times. This loop will be a continuous curve. In order to describe $\gamma$ more precisely we draw the lines $A\vee B$, $B\vee C$, $C\vee A$ in full. In this way the plane is divided into 7 compartments, one of them the triangle $\mathrm{△}(ABC)$.
Step 2
The loop $\gamma$ traverses the 6 unbounded compartments, and at each intersection with one of the above lines it has a corner. Within a given compartment $\gamma$ is an arc of an ellipse, whereby two of the three vertices A, B, C act as foci. The points A, B are foci for two such arcs. One of these arcs belongs to an ellipse with major axis $\ell -|AB|$ and the other to an ellipse with major axis $\ell -|BC|-|CA|$.