$\text { Friction on } 3 \mathrm{kg} \quad \mathrm{F}_{2}=0.3 \times 5 \times 10=15 \mathrm{N}$

$\begin{aligned} \text { acco. of system } a &=\frac{F}{\mathrm{m}} \\ &=\frac{10}{12}=\frac{5}{6} \end{aligned}$

Pseudo on $2 \mathrm{kg}$        $\mathrm{F}^{\prime}=2 \times \frac{5}{6}=\frac{5}{3} \mathrm{N}$

$\text { Friction on } 2 \mathrm{kg} \quad \mathrm{F}_{1}=0.2 \times 2 \times 10=4 \mathrm{N}$

So all blocks move together by $\frac{5}{6} \mathrm{m} / \mathrm{s}^{2}$