For the arrangement shown in the figure, mass of  A is 50 kg, mass of B  is 70 kg, coefficient of static friction of all plane surfaces is  ${\mu }_s=0.3$. Find the largest value of mass  C (in kg) so that blocks  A and  B remain at rest. (Neglect friction in the pulleys)

Let  $f_1$ and  $f_2$ be the friction force between A and B and between B and horizontal surface respectively. Limiting values if these frictional forces will be

$$\begin{gathered} f_{L_1}=0.3\times 50\times 10=150N \\ {f}_{L_2}=0.3\times 120\times 10=360N \end{gathered}$$
The FBD of the three blocks are shown in figure
Let  m is the mass of block C.  For A and B to remain at rest, block C should also remain at rest
For block C : $T=mg$  
For block  A :  $T_1=f_1$ 
For block  B :  $T=f_1+f_2+2T_1$
For largest value of mass of  C, the friction  $f_2$  must be towards left
$$\begin{gathered} T=3f_1+f_2=mg \\ {f}_1\le f_{L1}and f_2\le f_{L2} \end{gathered}$$

Hence  $mg\le 3f_{L1}+f_{L2}orm\le 81$
$\therefore$  maximum value of  $m=81kg$