Answer
\begin{align*}
\mathbf{T}& =\left(\frac{\cos t-t \sin t}{t+1}\right) \mathbf{i}+\left(\frac{\sin t+t \cos t}{t+1}\right) \mathbf{j}+\left(\frac{\sqrt{2} t^{1 / 2}}{t+1}\right) \mathbf{k}\\
\text{Length }& =\frac{\pi^{2}}{2}+\pi
\end{align*}
Work Step by Step
Since
$$\mathbf{r}=(t \cos t) \mathbf{i}+(t \sin t) \mathbf{j}+\frac{2 \sqrt{2}}{3} t^{3 / 2} \mathbf{k} $$
Then
$$ \mathbf{v}=(\cos t-t \sin t) \mathbf{i}+(\sin t+t \cos t) \mathbf{j}+\left(\sqrt{2} t^{1 / 2}\right) \mathbf{k}$$
and
\begin{aligned}
|v|&=\sqrt{(\cos t-t \sin t)^{2}+(\sin t+t \cos t)^{2}+(\sqrt{2} t)^{2}}\\
&=\sqrt{1+t^{2}+2 t}\\
&=\sqrt{(t+1)^{2}}=|t+1|=t+1
\end{aligned}
Hence
\begin{align*}
\mathbf{T}&=\frac{v}{|v|}\\
&=\left(\frac{\cos t-t \sin t}{t+1}\right) \mathbf{i}+\left(\frac{\sin t+t \cos t}{t+1}\right) \mathbf{j}+\left(\frac{\sqrt{2} t^{1 / 2}}{t+1}\right) \mathbf{k}\\
\text{Length }&=\int_{0}^{\pi}(t+1) d t\\
&=\left[\frac{t^{2}}{2}+t\right]_{0}^{\pi}\\
&=\frac{\pi^{2}}{2}+\pi
\end{align*}
