In the balanced chemical reaction${\text{IO}}_3^{-}+a{\text{I}}^{-}+b{\text{H}}^{+}\to c{\text{H}}_2\text{O}+d{\text{I}}_2$$a,b,c$ and $d$ respectively corresponds to:

Step-by-step balancing in acidic medium:Reduction half: $2{\text{IO}}_3^{-}+12{\text{H}}^{+}+10e^{-}\to {\text{I}}_2+6{\text{H}}_2\text{O}$ Oxidation half: $2{\text{I}}^{-}\to {\text{I}}_2+2e^{-}$ Multiply oxidation half by 5: $10{\text{I}}^{-}\to 5{\text{I}}_2+10e^{-}$ Add both: $2{\text{IO}}_3^{-}+10{\text{I}}^{-}+12{\text{H}}^{+}\to 6{\text{I}}_2+6{\text{H}}_2\text{O}$ Divide by 2: ${\text{IO}}_3^{-}+5{\text{I}}^{-}+6{\text{H}}^{+}\to 3{\text{I}}_2+3{\text{H}}_2\text{O}$ Comparing coefficients: $a=5,b=6,c=3,d=3$.