$\begin{align*}
A \cap C &= A \\
P(A) &= \frac{4\cdot5\cdot6\cdot6}{15\cdot14\cdot13} = \frac{24}{7\cdot13} \\
P(C) &= \frac{4\cdot11+5\cdot10+6\cdot9}{15\cdot14} = \frac{74}{15\cdot7} \\
P(A|C) &= \frac{P(A \cap C)}{P(C)} = \frac{P(A)}{P(C)} = \frac{\frac{24}{7\cdot13}}{\frac{74}{15\cdot7}} = \frac{12\cdot15}{37\cdot13}
\end{align*}$