On balancing the given redox reaction, the coefficients $a,b$ and $c$ are found to be, respectively:
$$a{\text{Cr}}_2{\text{O}}_7^{2-}+b{\text{SO}}_3^{2-}+c{\text{H}}^{+}\to 2{\text{Cr}}^{3+}+3{\text{SO}}_4^{2-}+4{\text{H}}_2\text{O}$$

Let's balance using the ion-electron method:Oxidation: ${\text{SO}}_3^{2-}+{\text{H}}_2\text{O}\to {\text{SO}}_4^{2-}+2{\text{H}}^{+}+2e^{-}$ Reduction: ${\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}+6e^{-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$ Multiply the oxidation half-reaction by 3 to equate electrons:$3{\text{SO}}_3^{2-}+3{\text{H}}_2\text{O}\to 3{\text{SO}}_4^{2-}+6{\text{H}}^{+}+6e^{-}$ Adding the two:${\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}+3{\text{SO}}_3^{2-}+3{\text{H}}_2\text{O}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}+3{\text{SO}}_4^{2-}+6{\text{H}}^{+}$ Simplify ${\text{H}}^{+}$ and ${\text{H}}_2\text{O}$:$1{\text{Cr}}_2{\text{O}}_7^{2-}+3{\text{SO}}_3^{2-}+8{\text{H}}^{+}\to 2{\text{Cr}}^{3+}+3{\text{SO}}_4^{2-}+4{\text{H}}_2\text{O}$ Coefficients are $a=1,b=3,c=8$.