Single phase heater loads on 3 phase supply
I believe the diagram below should represent your circuit:



The currents in the two heaters, Iab and Ibc are shifted 120 degrees as reflected on the vector diagrams. To find the current in line 2, IL2, we need to apply the Kirchhoff's current law to node B, which just says that the sum of the currents flowing into a node should be equal to the sum of the currents flowing out. In our case: Iab=IL2+Ibc or IL2=Iab-Ibc. 

Since all these currents are vectors, we determine IL2 by subtracting vector Ibc from vector Iab, as shown on the first vector diagram. With Iab=Ibc=I and the angle between them 120 degrees, IL2=I*sqrt(3), which is just geometry.

If the loads were not equal, the magnitude of the currents Iab and Ibc would not be the same, but the angle between them would still be 120 degrees and we would still be able to find IL2, geometrically, as shown on the second vector diagram.

With just two loads, Heater-1 and Heater-2, the currents in line 1 and line 2 are equal to the currents in the corresponding loads, Heater-1 and Heater-2, which, in your example, are the same: Iab=Ibc=I(heater).

If you connect the third load, say, Heater-3 (identical to two other heaters), between L1 and L3, the currents in line 1 and line 3, IL1 and IL3, would increase to I*sqrt(3). We can determine these currents applying the same Kirchhoff's law to nodes A and C.