Explanation

(a) Three forces $F_{41}$, $F_{42}$ and $F_{43}$ acting on $\mathcal{Q}$ as shown resultant of $F_{41} + F_{43}$
$= {\sqrt{2F}}_{each} = \sqrt{2}\frac{1}{4\pi\varepsilon_{0}}\frac{\mathcal{Q}q}{d^{2}}$
Resultant on $\mathcal{Q}$ becomes zero only when $q$ charges are of negative nature.
$F_{42} = \frac{1}{4\pi\varepsilon_{0}}\ \ \frac{\mathcal{Q \times Q}}{\left( \sqrt{2d} \right)^{2}}$
$\Rightarrow \sqrt{2}\frac{dQ}{d^{2}} = \frac{\mathcal{Q \times}Q}{2d^{2}}$
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$\sqrt{2} \times q = \frac{\mathcal{Q \times Q}}{2}$
$\therefore q = \frac{\mathcal{Q}}{2\sqrt{2}}$
$or\ \ \ \frac{\mathcal{Q}}{q} = - 2\sqrt{2}$