Balancing Redox reaction : with 2 species being oxidized
Your best bet is to consider the entire reactant $\ce{CrI3}$ as the reducing agent.

The chromium (III) iodide starts out as that compound and gets oxidized to a combination of chromium (VI) (in the chromate) and iodine (VII) (in the periodate) with a ratio of chromium to iodine matching the original compound:

$\ce{CrI3 -&gt; Cr^{VI} + 3 I^{VII}}$

Clearly the originally neutral reactant has been oxidized to atoms having a combined oxidation state of $+27$, so we balance this reaction with $27$ electrons:

$\ce{CrI3 -&gt; Cr^{VI} + 3 I^{VII} + 27 e^-}$

Each chlorine molecule takes up two of the electrons so the redox ratio must be $\ce{2 CrI3 : 27 Cl2}$.  Put those in and balance the remaining compounds as you normally would.