In triangle ABC , the point P divides BC internally in the ratio $3: 4$ and Q divides CA internally in the ratio $5: 3$. If AP and BQ intersect in a point $G$, then $G$ divides $A P$ internally in the ratio

Given triangle ABC with point $P$ on $BC$ dividing it $3:4$ so $BP:PC = 3:4$ and point $Q$ on $CA$ dividing it $5:3$ so $CQ:QA = 5:3$, determine the ratio $AG:GP$ using Menelaus’ Theorem.

Apply Menelaus’ Theorem to $\triangle ACP$ with transversal $B-G-Q$:

$\frac{AQ}{QC} \cdot \frac{CB}{BP} \cdot \frac{PG}{GA} = 1$

Substitute the given ratios:

From $CQ:QA = 5:3$, $\frac{AQ}{QC} = \frac{3}{5}$

From $BP:PC = 3:4$, $BC = BP + PC = 3k + 4k = 7k$ for some $k$, so $\frac{CB}{BP} = \frac{7}{3}$

Thus,

$\left(\frac{3}{5}\right)\left(\frac{7}{3}\right)\frac{PG}{GA} = 1$

$\frac{7}{5} \cdot \frac{PG}{GA} = 1$

$\frac{PG}{GA} = \frac{5}{7}$

Since $\frac{AG}{GP} = \frac{7}{5}$, point $G$ divides $AP$ internally in the ratio $7:5$.

Final answer: $\boxed{C}$