Consider the figure given below :


Let consider $2 q$ at $C$ to the left of $q$ at distance $X$ form $q$. Force on $2 \mathrm{q}$ at $\mathrm{C}$ (left of $\mathrm{q}$ ) is in opposite direction.
So net force will be zero is magnitude is equal
So, $\quad \mathrm{F}_{\mathrm{CA}}+\mathrm{F}_{\mathrm{CB}}=0$
or, $\quad\left(\mathrm{F}_{\mathrm{CA}}\right)=\left(-\mathrm{F}_{\mathrm{CB}}\right)$
$\frac{\mathrm{q}_{\mathrm{C}} \mathrm{q}_{\mathrm{A}}}{\left(4 \pi \epsilon_0\right)\left(\mathrm{r}_{(\mathrm{CA})}^2\right)}=\frac{-\mathrm{q}_{\mathrm{C}} \mathrm{q}_{\mathrm{B}}}{\left(4 \pi \epsilon_0\right)\left(\mathrm{r}_{(\mathrm{CB})}^2\right)}$
As given that, $\left(\mathrm{q}_{\mathrm{A}}=\mathrm{q}\right),\left(\mathrm{q}_{\mathrm{B}}=-3 \mathrm{q}\right),\left(\mathrm{q}_{\mathrm{C}}=+2 \mathrm{q}\right)$
When force of repulsion on it due to $q$ is balanced by force of attraction on it due to $(-3 q)$, at B, where $\mathrm{AB}=\mathrm{d}$ Thus, force of attraction by $(-3 q)=$ Force of repulsion by $q$
$$
\begin{aligned}
\frac{2 q \times q}{4 \pi \varepsilon_0 x^2} &=\frac{2 q \times 3 q}{4 \pi \varepsilon_0(x+d)^2} \\
\frac{1}{x^2} &=\frac{3}{(x+d)^2} \\
\left(x+d^2\right) &=3 x^2 \\
x^2+d^2+2 x d &=3 x^2=2 x^2-d^2
\end{aligned}
$$
So, $2 \mathrm{x}^2-2 \mathrm{dx}-\mathrm{d}^2=0$
$$
x=-\frac{d}{2} \pm \frac{\sqrt{3} d}{2}
$$
Negative sign be between $\mathrm{q}$ and $-3 \mathrm{q}$ so not considerable.
So, $x=\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3})$ to the left of $\mathrm{q}$