If $$\tan\left(\frac{\pi}{9}\right), x, \tan\left(\frac{7\pi}{18}\right)$$ are in arithmetic progression and $$\tan\left(\frac{\pi}{9}\right), y, \tan\left(\frac{5\pi}{18}\right)$$ are also in arithmetic progression, then $$|x - 2y|$$ is equal to:

We have that $$\tan\left(\frac{\pi}{9}\right),\,x,\,\tan\left(\frac{7\pi}{18}\right)$$ form an arithmetic progression, so by the definition of an A.P. the middle term is the average of the other two. Hence

$$2x=\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{7\pi}{18}\right).$$

Dividing by 2 we obtain

$$x=\frac{\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{7\pi}{18}\right)}{2}.$$

In the same way, the numbers $$\tan\left(\frac{\pi}{9}\right),\,y,\,\tan\left(\frac{5\pi}{18}\right)$$ are in arithmetic progression, so

$$2y=\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{5\pi}{18}\right)$$

and therefore

$$y=\frac{\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{5\pi}{18}\right)}{2}.$$

To handle these expressions we set

$$t=\tan\left(\frac{\pi}{9}\right)=\tan 20^{\circ}.$$

The angle $$\frac{7\pi}{18}=70^{\circ}$$ is the complement of $$20^{\circ}$$, so

$$\tan\left(\frac{7\pi}{18}\right)=\tan 70^{\circ}=\cot 20^{\circ}=\frac1{\tan 20^{\circ}}=\frac1t.$$

Thus

$$x=\frac{t+\dfrac1t}{2}.$$

Next we need $$\tan\left(\frac{5\pi}{18}\right)=\tan 50^{\circ}.$$ Let us denote

$$p=\tan 50^{\circ}.$$

Because $$70^{\circ}=50^{\circ}+20^{\circ},$$ we can apply the tangent addition formula

$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\,\tan B}.$$

Taking $$A=50^{\circ}$$ and $$B=20^{\circ},$$ we have

$$\tan 70^{\circ}=\frac{\tan 50^{\circ}+\tan 20^{\circ}}{1-\tan 50^{\circ}\tan 20^{\circ}}.$$

Substituting the symbols just introduced, this becomes

$$\frac1t=\frac{p+t}{1-pt}.$$

Cross-multiplying gives

$$t(p+t)=1-pt.$$

Expanding and bringing all terms to one side we get

$$tp+t^{2}=1-pt,$$

$$tp+pt+t^{2}=1,$$

$$2pt+t^{2}=1.$$

Solving for $$p$$ we find

$$p=\frac{1-t^{2}}{2t}.$$

Now we can write $$y$$ solely in terms of $$t$$:

$$y=\frac{t+p}{2}=\frac{t+\dfrac{1-t^{2}}{2t}}{2}.$$

It is more convenient to work directly with $$2y$$, because $$|x-2y|$$ appears in the question. We have from earlier

$$2y=t+p=t+\frac{1-t^{2}}{2t}.$$

Next we compute the required difference. Starting from the expression for $$x$$ we write

$$x-2y=\frac{t+\dfrac1t}{2}-\left(t+\frac{1-t^{2}}{2t}\right).$$

To combine terms we bring everything to a common denominator $$2t$$:

$$x-2y=\frac{t^{2}+1}{2t}-\frac{2t^{2}+1-t^{2}}{2t}.$$

Simplifying the numerator of the second fraction gives

$$2t^{2}+1-t^{2}=t^{2}+1,$$

so that

$$x-2y=\frac{t^{2}+1}{2t}-\frac{t^{2}+1}{2t}=0.$$

Hence

$$|x-2y|=|0|=0.$$

Hence, the correct answer is Option C.