#### Answer

$$(0,5,24 \pi)$$

#### Work Step by Step

Let $P\left(t_{0}\right)$ denote the point. Then
$$\mathbf{v}=(5 \cos t) \mathbf{i}-(5 \sin t) \mathbf{j}+12 \mathbf{k}$$ and \begin{align*}
26 \pi&=\int_{0}^{t_{0}} \sqrt{25 \cos ^{2} t+25 \sin ^{2} t+144} d t\\
&=\int_{0}^{t_{0}} 13 d t\\
&=13 t_{0}\end{align*}
Then $ t_{0}=2 \pi,$ and the point is $$P(2 \pi)=(5 \sin 2 \pi, 5 \cos 2 \pi, 24 \pi)=(0,5,24 \pi)$$