Balancing complex haloform redox reaction
As you noted, you need to find the limiting reactant. This is most of the time done by taking the one with a 1:1 ratio of educt:product. 

So in this case, the $\ce{ROCH_3}$. So you take that once. Now you see immediately that you need 2 $\ce{Na}$ on the right side, so you can put a 2 in front of the $\ce{NaOH}$. You have two $\ce{X}$ on the left, but 4 on the right, so put a 2 in front of the left one. Now count all your atoms. Subtract them from each other:

You have $a * \ce{H}$ on the left and $b * \ce{H}$ on the right. if $a-b=0$ its balanced. Do that for every reactant and you're done.

There are of course more strictly formulated methods, but in my time I noticed that a straightforward approach with a little thinking can be applied most universally without having to memorize some formulas.