Using whole molecules as oxidizing or reducing agents is my favorite method for dealing with redox reactions where atomic oxidation numbers are not straightforward. Here we use ethanol as a reducing reagent. With that let us design our half reactions:
Oxidation of ethanol to formate ion
We suppose ethanol is oxidized to give formate ion according t
$\ce{C2H5OH\to 2 HCO2^-}$
where we have balanced the carbon atoms. Introducing the water, hydroxide ions (in basic solution), and electrons we would have the fully balanced equation
$\ce{C2H5OH + (a)OH^-\to 2 HCO2^- + (b)H2O + (c)e^-}$
We expect the hydroxide ions to be on the left to balance all the negative charges on the right. To balance the hydrogen atoms we need $a+6=2b+2$, for oxygen we need $a+1=b+4$. Thus $b=7,a=10$ and when we put these results in the charge balance will give $(-10)=(-2)-c$ so $c=8$. Then
$\ce{C2H5OH +10 OH^-\to 2 HCO2^- + 7 H2O + 8 e^-}$
Reduction of chlorine
This is straightforward sincevinly chlorine atoms are reduced:
$\ce{Cl2 + 2e^-\to 2 Cl^-}$
Combine this with the oxidation half-reaction in the usual way and you have a balanced equation ... without the chloroform.
What happened?
We have a combination of two reactions, one oxidizing the alcohol to formate ion and the other oxidizing it to chloroform. We balanced just the oxidation to formate ion. Let's now look at the component where the alcohol is oxidized to chloroform.
Oxidation of ethanol to chloroform
Again we use ethanol as axwhole-molecule reducing agent. We must combine it with some chloride ions to balance carbon and chlorine, leading to an equation with the form
$\ce{C2H5OH + 6 Cl^- + (a)OH^-\to 2 CHCl3 + (b)H2O + (c)e^-}$
Balancing hydrogen requires $a+6=2b+2$ (again), and the oxygens give $a+1=b$. So $b=3,a=2$ and the charge balance then gives $(-6)-2=-c$ or $c=8$:
$\ce{C2H5OH + 6 Cl^- + 2 OH^-\to 2 CHCl3 + 3 H2O + 8 e^-}$
Combine this with the chlorine reduction given earlier and you have the balanced equation for the oxidation to chloroform. Note that in both component reactions each ethanol molecule gives off the same number of electrons and therefore should react with the same amount of chlorine. Thus the ratio of chlorine to ethanol is independent of how you combine these component reactions. (The actual value should emerge when you combine each of the oxidation half-reaction with the reduction.)
How to combine the two balanced equations
So you have two balanced equations, one with formate and one with chloroform. How do you combine them? Since any combination satisfies all the laws of physics, the answer is you can't with just pencil and paper. You have to do an experiment to determine the relative amounts of formate and chloroform you get, and match your combination of the two equations with those relative amounts. Don't forget to include your reaction conditions, as changing conditions could easily change the formate/chloroform ratio.