1. Write the skeletal reaction

   $a\,\mathrm{Cl_2} + b\,\mathrm{OH^-} \rightarrow c\,\mathrm{ClO^-} + d\,\mathrm{Cl^-} + e\,\mathrm{H_2O}$

2. Identify oxidation states of chlorine

   • In $\mathrm{Cl_2}$, oxidation state of Cl is $0$.
   • In $\mathrm{Cl^-}$, oxidation state of Cl is $-1$.
   • In $\mathrm{ClO^-}$, let oxidation state of Cl be $x$: $x + (-2) = -1 \Rightarrow x = +1$

   So chlorine undergoes disproportionation:

   • One Cl atom is oxidized: $0 \to +1$
   • One Cl atom is reduced: $0 \to -1$

3. Use the known disproportionation reaction in cold dilute alkali

   Chlorine in alkaline medium gives hypochlorite and chloride: $\mathrm{Cl_2 + 2OH^- \rightarrow ClO^- + Cl^- + H_2O}$

4. Verify atom balance

   For $\mathrm{Cl_2 + 2OH^- \rightarrow ClO^- + Cl^- + H_2O}$

   • Chlorine: LHS $=2$, RHS $=1+1=2$
   • Oxygen: LHS $=2$, RHS $=1+1=2$
   • Hydrogen: LHS $=2$, RHS $=2$
   • Charge: LHS $=-2$, RHS $=-1-1=-2$

   Hence it is balanced.

5. Extract coefficients

   Comparing with $a\,\mathrm{Cl_2} + b\,\mathrm{OH^-} \rightarrow c\,\mathrm{ClO^-} + d\,\mathrm{Cl^-} + e\,\mathrm{H_2O}$

   we get: $a=1,\quad b=2,\quad c=1,\quad d=1$

6. Match with options

   This corresponds to Option B.

Conclusion:

$\boxed{a,b,c,d = 1,2,1,1}$