5. B

Let the integer point be $\left(x_{0}, y_{0}\right)$, then its distance to the line $25 x-15 y+12=0$ is
$$d=\frac{\left|25 x_{0}-15 y_{0}+12\right|}{\sqrt{25^{2}+(-15)^{2}}}=\frac{\mid 25 x_{0}-15 y_{0}+12!}{5 \sqrt{34}}$$

Since $x_{0} 、 y_{0} \in \boldsymbol{Z}$, $25 . x_{0}-15 y_{0}$ is a multiple of 5.
Therefore, $\left|25 x_{0}-15 y_{0}+12\right| \geqslant 2$.
When $x_{0}=-1, y_{0}=-1$,
$$\left|25 x_{0}-15 y_{0}+12\right|=2 \text {. }$$

Thus, the minimum value sought is $\frac{\sqrt{34}}{85}$.