Answer
$$\left\{ {t:t > 0} \right\}{\text{ }}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \cos 2t{\bf{i}} + {e^{\sqrt t }}{\bf{j}} + \frac{{12}}{t}{\bf{k}} \cr
& {\bf{r}}\left( t \right) = \left\langle {\cos 2t,{e^{\sqrt t }},\frac{{12}}{t}} \right\rangle \cr
& {\text{the vector value is of the form }}{\bf{r}}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle \cr
& {\text{The domain of }}{\bf{r}}{\text{ is the largest set of values of }}t{\text{ on which all of }} \cr
& f\left( t \right){\text{, }}g\left( t \right){\text{ }}{\text{, and }}h\left( t \right){\text{ are defined}}{\text{. then }} \cr
& f\left( t \right) = \cos 2t{\text{, }}g\left( t \right) = {e^{\sqrt t }}{\text{ and }}g\left( t \right) = \frac{{12}}{t} \cr
& {\text{for }}f\left( t \right) = \cos 2t{\text{ the function is defined for all real numbers}}{\text{, then}} \cr
& {\text{the domain of }}f\left( t \right){\text{ is }}\left( { - \infty ,\infty } \right) \cr
& {\text{for }}g\left( t \right) = {e^{\sqrt t }}{\text{ the function is defined for }}t \geqslant 0,{\text{ then}} \cr
& {\text{the domain of }}g\left( t \right){\text{ is }}\left[ {0,\infty } \right) \cr
& {\text{for }}h\left( t \right) = \frac{{12}}{t}{\text{ the function is defined for }}t \ne 0,{\text{ then}} \cr
& {\text{the domain of }}h\left( t \right){\text{ is }}\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right) \cr
& {\text{The domain of }}{\bf{r}}\left( t \right){\text{ is the intersection of the domains of }}f\left( t \right){\text{, }}g\left( t \right){\text{ and }}h\left( t \right) \cr
& \left( { - \infty ,\infty } \right) \cap \left[ {0,\infty } \right) \cap \left[ {\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)} \right] = \left( {0,\infty } \right) \cr
& {\text{The domain is}}:{\text{ }}\left\{ {t:t > 0} \right\}{\text{ }} \cr} $$
