Answer
\begin{align*}
\mathbf{T} =\left(\frac{12}{13} \cos 2 t\right) \mathbf{i}-\left(\frac{12}{13} \sin 2 t\right) \mathbf{j}+\frac{5}{13} \mathbf{k}\\
\text{ Length } =13 \pi
\end{align*}
Work Step by Step
Since
$\mathbf{r}=(6 \sin 2 t) \mathbf{i}+(6 \cos 2 t) \mathbf{j}+5 t \mathbf{k} $
Then
$\mathbf{v}=(12 \cos 2 t) \mathbf{i}+(-12 \sin 2 t) \mathbf{j}+5 \mathbf{k}$
$ |\mathbf{v}|=\sqrt{(12 \cos 2 t)^{2}+(-12 \sin 2 t)^{2}+5^{2}}=13$
Hence
\begin{align*}
\mathbf{T}&=\frac{\mathbf{v}}{|\mathbf{v}|}\\
&=\left(\frac{12}{13} \cos 2 t\right) \mathbf{i}-\left(\frac{12}{13} \sin 2 t\right) \mathbf{j}+\frac{5}{13} \mathbf{k}\\
\text{ Length }&=\int_{0}^{\pi}|\mathbf{v}| d t\\
&=\int_{0}^{\pi} 13 d t=[13 t]_{0}^{\pi}\\
&=13 \pi
\end{align*}
