If $$ax^{2} + bx + 6$$ leaves the remainders 12 and 15 when divided by $$x - 3$$ and $$x + 4$$ respectively. then 7a + 14b =

A

$$\frac{9}{2}$$

B

$$\frac{17}{4}$$

C

$$\frac{27}{4}$$

D

$$\frac{11}{2}$$