Let $$A(a, b)$$, $$B(3, 4)$$ and $$(-6, -8)$$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $$P(2a + 3, 7b + 5)$$ from the line $$2x + 3y - 4 = 0$$ measured parallel to the line $$x - 2y - 1 = 0$$ is

A

$$\frac{15\sqrt{5}}{7}$$

B

$$\frac{17\sqrt{5}}{6}$$

C

$$\frac{17\sqrt{5}}{7}$$

D

$$\frac{\sqrt{5}}{17}$$