In triangle $$ABC$$, 

$$\tan A \colon \tan B \colon \tan C = 1 \colon 2 \colon 3$$. 
If $$\frac{AC}{AB} = \frac{p\sqrt{q}}{r}$$, where $$q$$ is Square free and $$\gcd(p, r) = 1$$ then the value of $$p + q + r$$ is

In any triangle $$\tan A + \tan B + \tan C = \tan A \tan B \tan C$$, so writing the tangents as $$t, 2t, 3t$$ gives $$6t = 6t^3$$ and $$t = 1$$. Hence $$\tan B = 2$$ and $$\tan C = 3$$, so $$\sin B = \frac{2}{\sqrt{5}}$$ and $$\sin C = \frac{3}{\sqrt{10}}$$. By the sine rule $$\frac{AC}{AB} = \frac{\sin B}{\sin C} = \frac{2\sqrt{2}}{3}$$, so $$p + q + r = 2 + 2 + 3 = 7$$.