Answer
There is a maximum at $~~x = 12.0~cm$
Work Step by Step
When $0 \leq x \leq 5.0~m$, the vertical components of the forces cancel out.
When $0 \lt x \leq 5.0~m$, the horizontal components of the forces are directed in the positive direction of the x axis. Therefore, the net force on particle 3 is greater than zero.
The maximum will occur at the point $x$ where the horizontal component of the force on particle 3 due to the other particles is a maximum.
We can find an expression for $F_x$ in terms of $x$:
$F_x = 2\times \frac{q_1~q_3}{4\pi~\epsilon_0~(\sqrt{x^2+d^2})^2}~cos~\theta$
$F_x = \frac{q_1~q_3}{2\pi~\epsilon_0~(x^2+d^2)}~\frac{x}{\sqrt{x^2+d^2}}$
$F_x = \frac{q_1~q_3}{2\pi~\epsilon_0}~\frac{x}{(x^2+d^2)^{3/2}}$
We can find $x$ when $F_x' = 0$:
$F_x' = \frac{q_1~q_3}{2\pi~\epsilon_0}~\frac{(x^2+d^2)^{3/2}-\frac{3}{2}(x^2+d^2)^{1/2}(2x^2)}{(x^2+d^2)^3} = 0$
$(x^2+d^2)^{3/2}-\frac{3}{2}(x^2+d^2)^{1/2}(2x^2) = 0$
$(x^2+d^2)^{3/2} = (3x^2)(x^2+d^2)^{1/2}$
$x^2+d^2 = 3x^2$
$2x^2 = d^2$
$x = \frac{d}{\sqrt{2}}$
$x = \frac{17.0~cm}{\sqrt{2}}$
$x = 12.0~cm$
There is a maximum at $~~x = 12.0~cm$
