$\begin{gathered}
  CBrC{l_3}\xrightarrow{{hv}}B{r^ \bullet }{ + ^ \bullet }CC{l_3} \hfill \\
  PhC{H_3}\xrightarrow{{ \bullet CC{l_3}}}CHC{l_3} + PhC{H_2\bullet},\xrightarrow{{CBrC{l_3}}}PhC{H_2}Br + \,\,\bullet CC{l_3} \hfill \\ 
\end{gathered}$, etc

Attack by the free radical on toluene occurs at the methyl side chain and not  in the ring because the $C - H$ bond in $Me$ is weaker than that of a ring- hydrogen atom and the benzyl free radical is far more stable than an aryl free  radical. 
The other point that requires explanation is why toluene is attacked by the  $CCl_3$ free radical and not by the bromine free radical. Activation energies  involving free radicals are usually very low and so the controlling factor is the  heat of reaction (or, more correctly, the free energy of reaction). The more  exothermic the reaction (greater is $\Delta G$), the more favoured is that  reaction. If the bromine atom attacks, the result is $HBr$ , the bond of which is  much weaker than the $C - H$ bond formed when $^\bullet CCl_3$ attacks. Hence, reaction
proceeds by the later route.