Let the plane $$P: 8x + \alpha_1 y + \alpha_2 z + 12 = 0$$ be parallel to the line $$L: \dfrac{x+2}{2} = \dfrac{y-3}{3} = \dfrac{z+4}{5}$$. If the intercept of $$P$$ on the y-axis is 1, then the distance between $$P$$ and $$L$$ is

A

$$\sqrt{\dfrac{2}{7}}$$

B

$$\dfrac{6}{\sqrt{14}}$$

C

$$\sqrt{\dfrac{7}{2}}$$

D

$$\sqrt{14}$$