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,

$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$

We are given that the ratios of the tangents of the three angles are

$$\tan A:\tan B:\tan C=1:2:3.$$

Using the ratio, we can write them as

$$\tan A=t,\qquad \tan B=2t,\qquad \tan C=3t.$$

Substituting these values in the triangle identity,

$$t+2t+3t=t(2t)(3t)$$

$$6t=6t^3$$

Dividing both sides by $$6t$$,

$$1=t^2$$

Since the angles of a triangle are positive, their tangents are positive in this case. Hence,

$$t=1.$$

Therefore,

$$\tan A=1,\qquad \tan B=2,\qquad \tan C=3.$$

$$\tan B=\dfrac{\text{opposite}}{\text{adjacent}}=2=\dfrac{2}{1},$$

we can consider a right triangle with sides $$2,1,\sqrt{5}$$. Therefore,

$$\sin B=\dfrac{2}{\sqrt{5}}.$$

Similarly,

$$\tan C=3=\dfrac{3}{1},$$

so the hypotenuse is

$$\sqrt{3^2+1^2}=\sqrt{10}.$$

Therefore,

$$\sin C=\dfrac{3}{\sqrt{10}}.$$

By the sine rule,

$$\dfrac{AC}{AB}=\dfrac{\sin B}{\sin C}.$$

Substituting the values of $$\sin B$$ and $$\sin C$$,

$$\dfrac{AC}{AB}=\dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{3}{\sqrt{10}}}.$$

Dividing by a fraction is the same as multiplying by its reciprocal:

$$\dfrac{AC}{AB}=\dfrac{2}{\sqrt{5}}\times\dfrac{\sqrt{10}}{3}.$$

Therefore,

$$\dfrac{AC}{AB} =\dfrac{2\sqrt{10}}{3\sqrt{5}}.$$

Since

$$\dfrac{\sqrt{10}}{\sqrt{5}}=\sqrt{\dfrac{10}{5}}=\sqrt{2},$$

we get

$$\dfrac{AC}{AB}=\dfrac{2\sqrt{2}}{3}.$$

Thus,

$${\dfrac{AC}{AB}=\dfrac{2\sqrt{2}}{3}}.$$

$$p=2,\qquad q=2,\qquad r=3,$$

then

$$p+q+r=2+2+3=7$$