Answer
$$P(-\pi)=(12 \sin (-\pi),-12 \cos (-\pi),-5 \pi)=(0,12,-5 \pi)$$
Work Step by Step
Let $P\left(t_{0}\right)$ denote the point. Then $$\mathbf{v}=(12 \cos t) \mathbf{i}+(12 \sin t) \mathbf{j}+5 \mathbf{k}$$ and
\begin{align*}
-13 \pi&=\int_{0}^{t_{0}} \sqrt{144 \cos ^{2} t+144 \sin ^{2} t+25} d t\\
&=\int_{0}^{t_{0}} 13 d t\\
&=13 t_{0}
\end{align*}
Then $ t_{0}=-\pi,$ and the point is
$$P(-\pi)=(12 \sin (-\pi),-12 \cos (-\pi),-5 \pi)=(0,12,-5 \pi)$$
