Less than perfect

We have an isosceles triangle of length a for equal sides.

$\text{Length of the hypotenuse =} \sqrt{a^2+a^2} = \sqrt{2}a$

$\text{Since the net potential energy of the system is 0 we have :}$

$\begin{aligned} \frac{1}{4\pi\epsilon_0}(\frac{q^2}{a}+\frac{qQ}{a}+\frac{qQ}{\sqrt{2}a}) = 0 \end{aligned}$

$\begin{aligned} \frac{q}{a} = -(\frac{qQ}{a}+\frac{qQ}{\sqrt{2}a}) \end{aligned}$

$\begin{aligned} 1 = -Q-\frac{Q}{\sqrt2} \end{aligned}$

$\text{Thus the answer is :}\begin{aligned} \color{#624F41}{\mathbf{Q}} = \frac{-2q}{2+\sqrt{2}} \end{aligned}$