For the reaction 
 $I^{-}+Cl{O}_3^{-}+H_{2}S{O}_4\to C{l}^{-}+HS{O}_4^{-}+I_2$
The correct statement(s) in the balanced equation is/are

Balancing the chemical equation by half-reaction method.
 $I^{-}+Cl{O}_3^{-}+H_{2}S{O}_4\stackrel{ }{\to }C{l}^{-}+HS{O}_4^{-}+I_2$
 $2I^{-}\stackrel{ }{\to }{I}_2+2e^{-}\times 3$
 $\frac{Cl{O}_3^{-}+6H^{+}+6e^{-}\stackrel{ }{\to }C{l}^{-}+3H_{2}O}{Cl{O}_3^{-}+6I^{-}+6H^{+}\stackrel{ }{\to }3I_2+C{l}^{-}+3H_{2}O}$
Adding  $6HS{O}_4^{-}$  to both sides.
 $Cl{O}_3^{-}+6I^{-}+6H_{2}S{O}_4\stackrel{ }{\to }3I_2+C{l}^{-}+6HS{O}_4^{-}+3H_{2}O$