If force F is increasing with time and at t=0, F=0, where will slipping first start?

$f_1\to$ force of friction between 2kg and 3kg

$$\begin{gathered} {\left(f_1\right)}_{\max }=0.5\times 3\times 10 \\ =15 \mathrm{N} \end{gathered}$$

$f_2\to$ force of friction between 2kg and 1kg

${\left(f_2\right)}_{\max }=0.3\times 5\times 10=15 \mathrm{N}$

$f_3\to$ force of friction between 1kg and ground

$$\begin{gathered} {\left(f_3\right)}_{\max }=0.1\times 6\times 10 \\ =6 \mathrm{N} \end{gathered}$$

When $F>6 \mathrm{N}$ system will start moving with a common acceleration

$a=\frac{F-6}{3+2+1}=\left(\frac{F}{6}-1\right)\mathrm{m}/{\mathrm{s}}^2$

$$\begin{gathered} f_1-6=\left(3\right)a=\frac{F}{2}-3 \\ {f}_1=\left(6+\frac{F}{2}-3\right) \\ =\left(\frac{F}{2}+3\right) \end{gathered}$$

Since, F is slightly greater than 6N

$$\begin{gathered} \therefore f_1<15 \mathrm{N} \\ \text{ or } <{\left(f_1\right)}_{\max } \end{gathered}$$ 

$\therefore$ No slipping will occur here

$$\begin{gathered} f_2-6=\left(1\right)\left(a\right)=\frac{F}{6}-1 \\ {f}_2=\frac{F}{6}+5 \end{gathered}$$

Again $f_2<{\left(f_2\right)}_{\max }$. So no slip will take place here also.