To balance the given redox reaction, we need to follow these steps:$\\$1. Identify the oxidation and reduction half-reactions.$\\$• $$Reduction: $Cr_2O_7^{2-} \rightarrow Cr^{3+} \\$• $$Oxidation: $SO_3^{2-} \rightarrow SO_4^{2-} \\$2. Balance each half-reaction separately.$\\$Reduction half-reaction:$\\ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \\$Oxidation half-reaction:$\\ SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^- \\$3. Equalize the number of electrons transferred in each half-reaction.$\\$• Multiply the oxidation half-reaction by 3:$\\ 3(SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-) \\$This gives:$\\ 3SO_3^{2-} + 3H_2O \rightarrow 3SO_4^{2-} + 6H^+ + 6e^- \\$4. Add the balanced half-reactions together.$\\ Cr_2O_7^{2-} + 14H^+ + 6e^- + 3SO_3^{2-} + 3H_2O \rightarrow 2Cr^{3+} + 7H_2O + 3SO_4^{2-} + 6H^+ + 6e^- \\$5. Simplify the equation by canceling out common terms.$\\$• Cancel $6e^-$ from both sides.$\\$• Cancel $6H^+$ from both sides.$\\$• Simplify $3H_2O$ and $7H_2O$ to $4H_2O. \\$The balanced equation is:$\\ Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O \\$6. Identify the coefficients:$\\$• $a = 1 \\$• $b = 3 \\$• $c = 8 \\$Therefore, the coefficients are $1, 3, 8$, which corresponds to Option 1.