Process of a substitution reaction
This chemistry was first described in this 1960 JACS paper here.

The first step is cleavage of the $\ce{Br-CCl3}$ bond to give $\ce{Br^.}$ and $\ce{^.CCl3}$.
The second step is $\ce{Br^.}$ abstracting a proton from toluene to give $\ce{HBr}$ and a benzyl radical.
The third step is the $\ce{^.CCl3}$ radical abstracting a proton from toluene to give chloroform and a benzyl radical.
The fourth step is a benzyl radical abstracting $\ce{Br}$ from $\ce{Br-CCl3}$ to give benzyl bromide and $\ce{^.CCl3}$.
The fifth step is two $\ce{^.CCl3}$ radicals quenching to give hexachloroethane $\ce{Cl3C-CCl3}$.

So overall the products are $\ce{HBr}$, Benzyl bromide, Chloroform, and Hexachloroethane.
Reactions 3 and 4 are the major portion of the reaction so equimolar amounts of benzyl bromide and chloroform are formed along with small amounts of $\ce{HBr}$ and hexachloroethane (approx 20:1).