Three large plates are arranged as shown. How much charge will flow through the key 'K' if it is closed?

Before closing the switch

After closing the switch

$\begin{array}{l}{\mathrm{q}}_1+{\mathrm{q}}_2=2\mathrm{Q}\\ {\mathrm{V}}_{\mathrm{AB}}={\mathrm{V}}_{\mathrm{AC}}\text{ or }\frac{{\mathrm{q}}_1}{{\mathrm{C}}_1}=\frac{{\mathrm{q}}_2}{{\mathrm{C}}_2}\\ \frac{{\mathrm{q}}_1}{{\mathrm{q}}_2}=\frac{{\mathrm{C}}_1}{{\mathrm{C}}_2}\text{ or }{\mathrm{q}}_1=2{\mathrm{q}}_2\end{array}$
Solving, we get $q_1=\frac{4Q}{3},q_2=\frac{2Q}{3}$
Charge flown through K is $-\frac{Q}{2}-\left(-q_1\right)=-\frac{Q}{2}+\frac{4Q}{3}=5Q/6$