Answer
$$\left\{ {t:t \ne 1,t \ne 2} \right\}{\text{ }}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \frac{2}{{t - 1}}{\bf{i}} + \frac{3}{{t + 2}}{\bf{j}} \cr
& {\bf{r}}\left( t \right) = \left\langle {\frac{2}{{t - 1}},\frac{3}{{t + 2}}} \right\rangle \cr
& {\text{the vector value is of the form }}{\bf{r}}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle \cr
& {\text{The domain of }}{\bf{r}}{\text{ is the largest set of values of }}t{\text{ on which all of }} \cr
& f\left( t \right){\text{, }}g\left( t \right){\text{ }}{\text{, and }}h\left( t \right){\text{ are defined}}{\text{. then }} \cr
& f\left( t \right) = \frac{2}{{t - 1}}{\text{ and }}g\left( t \right) = \frac{3}{{t + 2}} \cr
& {\text{for }}f\left( t \right){\text{ the denominator cannot be }}0.{\text{ then }}t - 1 \ne 0,\,\,\,\,t \ne 1 \cr
& {\text{the domain of }}f\left( t \right){\text{ is all real numbers }}\Re \ne 1 \cr
& {\text{for }}h\left( t \right){\text{ the denominator cannot be }}0.{\text{ then }}t + 2 \ne 0,\,\,\,\,t \ne - 2 \cr
& {\text{The domain of }}{\bf{r}}\left( t \right){\text{ is the intersection of the domains of }}f\left( t \right){\text{ and }}h\left( t \right) \cr
& {\text{The domain is}}:{\text{ }}\left\{ {t:t \ne 1,t \ne 2} \right\}{\text{ }} \cr} $$
