Distance between centre to centre between columns is $2.3$. Assume width of footing as $1.5m$ and S.B.C of soil as $150 kN/m^2$. Also draw a neat sketch showing reinforcement details.

Area of footing $= \dfrac {(P_A+P_B )+10% \text { extra as self wt.}}{\text {SBC}}\\ =\dfrac{(850+1150)+\frac {10}{100}×(850+1150)}{150 } \\ A_f=14.67 m^2 \\ Now A_f=L_f×B_f \\ 14.67= L_f×1.5 \\ L_f=9.78 \\ \text { Find C.G. of load Take moment of loads @ ‘A’ }\\ x=\dfrac {[(1150×2.3)+(850×0)]}{(1150+850)}=1.325 m \\ \text { Factored upward soil pressure }(w_d) \\ w_d=\dfrac {1.5×[850+1150]}{9.78×1.5}=204.5 kN/m^2\\ \text { Upward soil pressure @ G of longitudinal axis. } \\ = w_d×B_f=204.5×1.5=306.75 kN/m $

$$B.M @ a =306.75 ×\dfrac {3.57^2}2=1957.74 kNm$$

$B.M @ b =306.75 ×\dfrac {3.915^2}2=2350.81 kNm \\ \dfrac x{179.9}=\dfrac {2.3-x}{5.25×625} \\ x= 0.586 m \\ B.M_{max}=1902 kNm \\ \text {Two way shear: [Check for big column only] } \\ M_{umax}=0.138f_{ck} bd^2 \\ 2350.81×10^6=0.138×20×1500×dr^2 \\ dr=753.54 mm \\ Provide d=1000mm$

$$\text {Net shear force} = P_u-[(\text {Area of critical section })×wd] $$

$ =1725-[(1.7×1.23)×204.5] \\ =1297.39 kN \\ \text { Area resisting = Perimeter of critical section } ×d \\ =[2 ×(1.7+2.3)]×1=5.86 m^2 \\ Z_v=\dfrac {1297.39×10^3}{5.86×10^6 }=0.22 N/mm^2 \\ Z_c \space \space permissible=Z_c'×k_s \\ Z_c'=0.25\sqrt {f_{ck}}=0.25\sqrt{20}=1.12 N/mm^2 \\ k_s=0.5+\dfrac {0.23}{0.7}=0.828 \lt 1=0.828 \\ Z_c permi=1.12×0.828=0.927N/mm^2 \gt 0.22 N/mm^2 \\ \text { Column A: }\\ \text { Band Width: } (b+2d)=[500 +2×1000]=2500 mm $

$$M_u=(2.5×0.635×204.5)×\dfrac {0.635}2$$

$M_u=103.07 kNm \\ Ast =(\dfrac {0.5×20×2500×1000}{415})×[1-\sqrt{1-\dfrac {4.6×103.07×10^6}{20×2500×1000^2 }}]\\ Astp =286.3 mm^2 \\ Astmin=\dfrac {0.12}{100} bD=\dfrac {0.12}{100} ×2500×(1000+100)=3300 mm^2 \\ Provide 3300 mm^2 Ast \\ \text { Hence provide 17-16 mm∅ bars }\\ Astp = 3418.05 mm^2 \\ Column B:-\\ \text {Band Width }= b+2d = 700+2×1000=2700 mm $

$$M_u=2.7×0.635×204.5×\dfrac {0.635}2=111.32 kNm$$

$Ast =\dfrac {0.5×20×2700×1000}{415}×[1-\sqrt{1-\dfrac {4.6×111.32×10^6}{20×2700×1000^2 }}] \\ Ast=309.21 mm^2 \\ Ast_{min}=\dfrac {0.12}{100}×2700×1000=3240 mm^2 \\ \text { Take Ast as } 3240 mm^2 \\ \text {Provide same as 17 -16 mm ∅ bars }\\ Astp=3418.05 mm^2$