Answer
The charge of particle 3 is $~~-5.7~\mu C$
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 see that $U=0$ when the x coordinate of particle 3 is $x = 10~cm$
We can find $q_3$, the charge of particle 3:
$U = \frac{1}{4\pi~\epsilon_0}~(\frac{q_1~q_2}{4.0}+\frac{q_1~q_3}{14}+\frac{q_2~q_3}{10}) = 0$
$\frac{q_1~q_2}{4.0}+\frac{q_1~q_3}{14}+\frac{q_2~q_3}{10} = 0$
$35q_1~q_2+10q_1~q_3+14q_2~q_3 = 0$
$q_3~(10q_1+14q_2) = -35q_1~q_2$
$q_3 = -\frac{35q_1~q_2}{10q_1+14q_2}$
$q_3 = -\frac{(35)(5.0~\mu C)(3.0~\mu C)}{(10)(5.0~\mu C)+(14)(3.0~\mu C)}$
$q_3 = -5.7~\mu C$
The charge of particle 3 is $~~-5.7~\mu C$.