Answer :
The electric field exerted by a point charge is given by:
[tex]E=k\frac{q}{r^2}[/tex]where q is the charge and r is the distance to the point where we want to calculate the electric field.
In this case we have three charges with the following properties:
[tex]\begin{gathered} q_1=6\times10^{-9},\text{ }r_1=0.0844 \\ q_2=-4\times10^{-9},\text{ }r_2=0.0312 \\ q=-10\times10^{-9},\text{ }r \end{gathered}[/tex]Now, the total electric field on a point is the addition of all the charges; in this case we want the net field to be zero. Then we have:
[tex]\begin{gathered} k\frac{6\times10^{-9}}{(0.0844)^2}-k\frac{4\times10^{-9}}{(0.0312)^2}-k\frac{10\times10^{-9}}{r^2}=0 \\ \frac{10}{r^2}=\frac{6}{(0.0844)^2}-\frac{4}{(0.0312)^2} \\ \frac{10}{r^2}=-3266.84 \\ r^2=-\frac{10}{3266.84} \end{gathered}[/tex]Now, since the last equation does not have a real solution this means that this distribution of charges will not exert a zero electric field on the origin.
Therefore, there's no possible distance for the field to be zero in this charge configuration
