Answer
$\textbf{T}(1) = \langle\frac{2}{\sqrt 5},0,\frac{-1}{\sqrt 5}\rangle$
Work Step by Step
$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$
$\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$
$\textbf{r}(t) = \langle 6t,6,3/t\rangle$
$\textbf{r}'(t) = \langle 6,0,-3/t^2\rangle$
$|\textbf{r}'(t)| = \sqrt {(6)^2 + (0)^2 + (-3/t^2)^2} = \sqrt {36 +9/t^4}$
$\textbf{T}(t) = \frac{\textbf{r}'(t)}{|\textbf{r}'(t)|} = \frac{\langle 6,0,-3/t^2\rangle}{ \sqrt {36 +9/t^4}}$
$\textbf{T}(1) = \frac{\langle 6,0,-3/(1)^2\rangle}{ \sqrt {36 +9/(1)^4}} = \langle\frac{2}{\sqrt 5},0,\frac{-1}{\sqrt 5}\rangle$
