Three charges Q, $+q$ and $+q$ are placed at the vertices of the right angle triangle (isosceles triangle) as shown. The net electrostatic energy of the configuration is zero, if Q is equal to [a $=$ length of two equal sides of a triangle]

The net electrostatic potential energy of a system of charges is the sum of the potential energies of all pairs of charges.

Let the charges be placed at the vertices of the right-angled isosceles triangle. From the given figure, the charges are $Q$, $+q$, and $+q$.
The distances between the pairs of charges are:
1. Distance between $Q$ and $+q$ (vertical side) $= a$
2. Distance between $+q$ and $+q$ (horizontal side) $= a$
3. Distance between $Q$ and $+q$ (hypotenuse) $= \sqrt{a^2 + a^2} = a\sqrt{2}$

The total electrostatic potential energy $U$ of the system is given by:
$U = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{Q \cdot q}{a} + \dfrac{q \cdot q}{a} + \dfrac{Q \cdot q}{a\sqrt{2}} \right)$

Given that the net electrostatic energy of the configuration is zero, we have $U = 0$:
$\dfrac{1}{4\pi\epsilon_0} \left( \dfrac{Qq}{a} + \dfrac{q^2}{a} + \dfrac{Qq}{a\sqrt{2}} \right) = 0$

Dividing the entire equation by $\dfrac{q}{4\pi\epsilon_0 a}$ (assuming $q \neq 0$), we get:
$Q + q + \dfrac{Q}{\sqrt{2}} = 0$

Grouping the terms with $Q$:
$Q \left( 1 + \dfrac{1}{\sqrt{2}} \right) + q = 0$

$Q \left( \dfrac{\sqrt{2} + 1}{\sqrt{2}} \right) = -q$

$Q = \dfrac{-\sqrt{2}q}{\sqrt{2} + 1}$

Answer: $\dfrac{-\sqrt{2}q}{1+\sqrt{2}}$