Two large metal plates are originally oriented horizontally and separated by a distance of $3d$. A conducting wire joins them and initially plates are uncharged.

A third identical plate with charge $Q$ is put parallel to them at a distance of $d$ from upper plate.

Then,

Center plates make an arrangement equivalent to system shown in figure.

So we have two capacitors with same potential difference $V$ and total charge Q.

Also, $C_1=\frac{{\epsilon }_{0}A}{d},C_2=\frac{{\epsilon }_{0}A}{2d}$

$Q_1+Q_2=Q$

and $V_1=V_2=V$

Then, $\frac{Q_1}{C_1}=\frac{Q_2}{C_2}\Rightarrow \frac{Q_{1}d}{{\epsilon }_{0}A}=\frac{Q_{2}2d}{{\epsilon }_{0}A}$

$Q_1=2Q_2;2Q_2+Q_2=Q$

So, $Q_2=\frac{Q}{3}$

$\Rightarrow \text{Charge on lower plate is }-\frac{Q}{3}$

and $Q_1=\frac{2Q}{3}, \mathrm{charge} \mathrm{on} \mathrm{upper} \mathrm{plate} \mathrm{is}-\frac{2Q}{3}$

Also, $\Delta V=\frac{Q_1}{C_1}=\frac{2Qd}{3{\epsilon }_{0}A}$