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The internal bisector of an angle $\text{A}$ in a triangle $\text{ABC}$ meets the side $\text{BC}$ at point $\text{D. AB = 4, AC = 3}$ and angle $\text{A} = 60^{\circ}$. Then what is the length of the bisector $\text{AD}?$

- $\frac{12 \sqrt{3}}{7}$
- $\frac{12 \sqrt{13}}{7}$
- $\frac{4 \sqrt{13}}{7}$
- $\frac{4 \sqrt{3}}{7}$

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