Explanation

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