Answer
There is no change in the electric potential energy of the three-particle system.
The work done by the net electric force on the particle of charge $+q$ due to the other two particles is $~~0$
The work done by our force on the particle of charge $+q$ is $~~0$
Work Step by Step
We can write a general expression for the electric potential energy of a system of charged particles:
$U = \frac{1}{4\pi~\epsilon_0}~\sum \frac{q_i~q_j}{r_{ij}}$
We can find an expression for the electric potential energy of the system when the $+q$ charge is at point B:
$U_B = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{d}+\frac{q~Q}{2d}+\frac{Q^2}{3d})$
We can find an expression for the electric potential energy of the system when the $+q$ charge is at point C:
$U_C = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{d}+\frac{q~Q}{2d}+\frac{Q^2}{3d})$
Note that $U_B = U_C$
There is no change in the electric potential energy of the three-particle system.
We can find the work done by the net electric force on the particle of charge $+q$ due to the other two particles:
$Work = -\Delta U = 0$
The work done by the net electric force on the particle of charge $+q$ due to the other two particles is $~~0$
We can find the work done by our force on the particle of charge $+q$:
$Work = \Delta U = 0$
The work done by our force on the particle of charge $+q$ is $~~0$
