Electric Field of Superposition of Line Charges to Form an Equilateral Triangle Question: If three line charges with charge densities $\rho_1 = -2\rho_2= -2\rho_3$, and all length $L$ form an equilateral triangle, what is the force felt by a charge $Q$ at the center of the triangle? 
Solution Attempt, and Specific Question: 
In solving this question I noticed that all three line charges must be an equal perpendicular distance from the center, so maybe by superposition I could reduce the problem to an equivalent single line charge with density some fraction of $\rho_1$. My specific question is whether $\rho_2$ and $\rho_3$ exactly cancel off $\rho_1$ so no force is felt by the charge. 
My reasoning for why an exact cancellation should occur is that if we consider not the electric field but the electric potential then the total potential is just a sum of scalar quantities. The electric potential along the perpendicular bisector of a line charge is dependent on the length of the line charge and the distance along the bisector. But line charges are all at an equal distance along their bisectors so the total potential is proportional to $$\rho_1(something) -\rho_1/2(something) -\rho_1/2(something)$$ which equals $$(\rho_1 -\rho_1/2 -\rho_1/2)(something)$$ = $$0$$. 
The potential is $0$ so the field and hence the force must be $0$. 
 A: Imagine a charged equilateral triangle with the same uniform charge density along each side.
The net electric field at the centre of the triangle would be zero but the potential would not if the zero of potential is taken to be at infinity.
Imagine bringing a unit positive test charge coming from from infinity to the centre of the triangle.
An external force would have to do work (positive if the charge density is positive and negative if the charge density is negative) in moving from infinity to the centre of the triangle and hence the potential at that point is non zero.
I think the confusion might arise because the electric field is equal to minus the potential gradient.
Zero electric field means that the potential gradient is zero so the potential is a constant but not necessarily zero.

The changes on the three sides of the triangle are $\rho_1L, -\rho_2L$ and  $-\rho_3L$ which rewritten in terms of charge density $\rho_1$ are $\rho_1L, -\frac{\rho_1 L}{2}$ and  $-\frac{\rho_1L}{2}$.
If all the changes were $-\frac{\rho_1L}{2}$ then by symmetry the electric field at the centre of the triangle would be zero.
You can use superposition by finding out what charge (density) needs to be added to $-\frac{\rho_1L}{2}$ to get $\rho_1L$.  
All you then need to do is to find the electric field due to that added charge (density).

Update in response to some comments.
I have tried to show diagrammatically how superposition works for this example.

If the line of charge with line charge density $\rho_1$ produces an electric field of magnitude $E$ the a line of charge with half the line charge density pr4oduces an electric field of magnitude $\frac E 2$.
Diagram 1 shows the initial problem.   
Diagram 2 shows the three equal line charges producing zero electric field.  
Diagram 3 shows the electric field which must be added to the arrangement of changes in diagram 2 to produce the charges and electric fields in diagram 1.
@DWin Your comment was correct.  
Looking at diagram 1 and adding the electric field vectorially a resultant electric field of magnitude
$E + \frac E 2 \; \cos 60^\circ + \frac E 2 \; \cos 60^\circ = \frac{3E}{2}$
