Answer
Net force on $Q_{1} \approx 116N$ to the left
Net force on $Q_{2}\approx 564N$ to the right
Net force on $Q_{3}\approx 448N$ to the left
Work Step by Step
Net force on $Q_{1}$ is
$$F_{1net} = F_{12} + F_{13}$$ where $F_{12}$ means the force on 1 by 2.
$F_{12} = k\frac{Q_{1}Q_{2}}{r_{12}^{ 2}}$, where Q is an absolute value.
$F_{12} = 9.0\times10^{9} Nm^{2}/C^{2}\frac{(65\times10^{-6}C)(48\times10^{-6}C)}{0.35m^{2}}$
$F_{12} = -229N$
Repeat for $F_{13}$.
$F_{13} = 113N$
Force on $Q_{1} = -229N + 113N \approx -116N$
The negative sign indicates the direction is to the left.
Repeat steps for force on $Q_{2}$.
We know that the magnitude of $F_{21} = F_{12}$, but the direction is opposite.
$F_{2net} = F_{21} + F{23}$
$F_{21} = 229N$
$F_{23} = 335N$
Force on $Q_{2} = 229N + 335N \approx 564N$
Now find force on $Q_{3}$.
$F_{3net} = F_{31} + F_{32}$
Force on $Q_{3} = -113N + 335N \approx -448N$
