Find the charge which will flow after switch is closed through sections 1,2 and 3in the directions indicator

When switch is open charge on each capacitor is  ${\mathrm{q}}_{\mathrm{o}}={\mathrm{C}}_{\mathrm{e}}\times {\mathrm{V}}_{\mathrm{e}}=\frac{3\times 2}{5}\times 120=144 \mathrm{\mu c}$

When switch is closed

q₁ = 120 $\mathrm{\mu }$C; q₂ = 180 $\mathrm{\mu }$C 

Charge through section 1 = q₁ -q₀ $=120-144= -24 \mathrm{\mu c}$

Charge through section 2 = –q₂ – (–q₀) $=-180 +144=-36 \mathrm{\mu c}$

Charge through section 3 = q₂ – q₁   $=180-120=60 \mathrm{\mu c}$