Answer
$$x = 1 - \sqrt 3 \pi t;{\text{ }}y = \sqrt 3 + \pi t,{\text{ }}z = 1 + 3t$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 2\cos \pi t{\bf{i}} + 2\sin \pi t{\bf{j}} + 3t{\bf{k}};{\text{ }}{t_0} = \frac{1}{3} \cr
& {\text{Calculate }}{\bf{r}}\left( {{t_0}} \right) \cr
& {\bf{r}}\left( {\frac{1}{3}} \right) = 2\cos \left( {\frac{\pi }{3}} \right){\bf{i}} + 2\sin \left( {\frac{\pi }{3}} \right){\bf{j}} + 3\left( {\frac{1}{3}} \right){\bf{k}} \cr
& {\bf{r}}\left( {\frac{1}{3}} \right) = 2\left( {\frac{1}{2}} \right){\bf{i}} + 2\left( {\frac{{\sqrt 3 }}{2}} \right){\bf{j}} + {\bf{k}} \cr
& {\bf{r}}\left( {\frac{1}{3}} \right) = {\bf{i}} + \sqrt 3 {\bf{j}} + {\bf{k}} \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( {{t_0}} \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {2\cos \pi t{\bf{i}} + 2\sin \pi t{\bf{j}} + 3t{\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = - 2\pi \sin \pi t{\bf{i}} + 2\pi \cos \pi t{\bf{j}} + 3{\bf{k}} \cr
& {\bf{r}}{\text{'}}\left( {\frac{1}{3}} \right) = - 2\pi \sin \left( {\frac{\pi }{3}} \right){\bf{i}} + 2\pi \cos \left( {\frac{\pi }{3}} \right){\bf{j}} + 3{\bf{k}} \cr
& {\bf{r}}{\text{'}}\left( {\frac{1}{3}} \right) = - 2\pi \left( {\frac{{\sqrt 3 }}{2}} \right){\bf{i}} + 2\pi \left( {\frac{1}{2}} \right){\bf{j}} + 3{\bf{k}} \cr
& {\bf{r}}{\text{'}}\left( {\frac{1}{3}} \right) = - \sqrt 3 \pi {\bf{i}} + \pi {\bf{j}} + 3{\bf{k}} \cr
& {\text{Let }}{{\bf{r}}_0} = {\bf{r}}\left( {{t_0}} \right){\text{ and }}{{\bf{v}}_0}{\text{ = }}{\bf{r}}{\text{'}}\left( {{t_0}} \right){\text{ }} \cr
& {\text{The tangent line to the graph of }}{\bf{r}}\left( t \right){\text{ at }}{{\bf{r}}_0}{\text{ is given by the vector }} \cr
& {\text{equation}} \cr
& {\bf{r}} = {{\bf{r}}_0} + t{{\bf{v}}_0} \cr
& {\bf{r}} = {\bf{r}}\left( {{t_0}} \right) + t{\bf{r}}{\text{'}}\left( {{t_0}} \right){\text{ }} \cr
& {\bf{r}} = {\bf{i}} + \sqrt 3 {\bf{j}} + {\bf{k}} + t\left( { - \sqrt 3 \pi {\bf{i}} + \pi {\bf{j}} + 3{\bf{k}}} \right) \cr
& {\bf{r}} = {\bf{i}} + \sqrt 3 {\bf{j}} + {\bf{k}} - \sqrt 3 \pi t{\bf{i}} + \pi t{\bf{j}} + 3t{\bf{k}} \cr
& {\bf{r}} = \left( {1 - \sqrt 3 \pi t} \right){\bf{i}} + \left( {\sqrt 3 + \pi t} \right){\bf{j}} + \left( {1 + 3t} \right){\bf{k}} \cr
& {\text{Thus}},{\text{the parametric equations of the tangent line at }}t = {t_0}{\text{ are:}} \cr
& x = 1 - \sqrt 3 \pi t;{\text{ }}y = \sqrt 3 + \pi t,{\text{ }}z = 1 + 3t \cr} $$
