jmc

By Pythagoras, triangle $ABC$ is right with $\angle B=90^{\circ }$. Then the area of triangle $ABC$ is $1/2\cdot AB\cdot BC=1/2\cdot 3\cdot 4=6$.

Since $G$ is the centroid of triangle $ABC$, the areas of triangles $BCG$, $CAG$, and $ABG$ are all one-third the area of triangle $ABC$, namely $6/3=2$.

We can view $PG$ as the height of triangle $BCG$ with respect to base $BC$. Then $$\frac{1}{2}\cdot GP\cdot BC=2,$$so $GP=4/BC=4/4=1$. Similarly, $GQ=4/AC=4/5$ and $GR=4/AB=4/3$. Therefore, $GP+GQ+GR=1+4/5+4/3=\boxed{\frac{47}{15}}$.