If the tangent to the curve $$y = \frac{x}{x^2 - 3}$$, $$x \in R$$, $$x \neq \pm\sqrt{3}$$, at a point $$(\alpha, \beta) \neq (0, 0)$$ on it is parallel to the line $$2x + 6y - 11 = 0$$, then:

We look at the curve $$y=\dfrac{x}{x^{2}-3}$$, where $$x\in\mathbb R,\;x\neq\pm\sqrt{3}$$. The slope of the tangent at any point is given by the derivative $$\dfrac{dy}{dx}$$.

Recalling the quotient rule, for $$y=\dfrac{u}{v}$$ we have $$\dfrac{dy}{dx}=\dfrac{v\,u'-u\,v'}{v^{2}}.$$ Here $$u=x,\;u'=1,\;v=x^{2}-3,\;v'=2x.$$ So

$$\dfrac{dy}{dx}=\dfrac{(x^{2}-3)(1)-x(2x)}{(x^{2}-3)^{2}} =\dfrac{x^{2}-3-2x^{2}}{(x^{2}-3)^{2}} =\dfrac{-x^{2}-3}{(x^{2}-3)^{2}} =-\,\dfrac{x^{2}+3}{(x^{2}-3)^{2}}.$$

The tangent is said to be parallel to the line $$2x+6y-11=0.$$ For any line in the form $$Ax+By+C=0,$$ its slope is $$m=-\dfrac{A}{B}.$$ Hence the slope of this line is $$m=-\dfrac{2}{6}=-\dfrac13.$$

Because parallel lines have equal slopes, we set

$$-\,\dfrac{x^{2}+3}{(x^{2}-3)^{2}}=-\dfrac13.$$

Both sides are negative, so cancelling the minus signs gives

$$\dfrac{x^{2}+3}{(x^{2}-3)^{2}}=\dfrac13.$$

Cross-multiplying yields

$$3(x^{2}+3)=(x^{2}-3)^{2}.$$

Letting $$t=x^{2}$$ simplifies the equation:

$$3(t+3)=(t-3)^{2} \;\Longrightarrow\;3t+9=t^{2}-6t+9.$$

Bringing all terms to one side:

$$0=t^{2}-6t+9-3t-9=t^{2}-9t \;\Longrightarrow\;t(t-9)=0.$$

Thus $$t=0\quad\text{or}\quad t=9.$$ Since $$t=x^{2}$$ and the point $$(\alpha,\beta)$$ is not $$(0,0),$$ we discard $$t=0.$$ Therefore $$x^{2}=9\;\Longrightarrow\;x=\pm3.$$ So $$\alpha=3\quad\text{or}\quad\alpha=-3.$$

Substituting into the original curve to find $$\beta:$$

For $$x=3:$$ $$\beta=y=\dfrac{3}{3^{2}-3}=\dfrac{3}{9-3}=\dfrac{3}{6}=\dfrac12.$$

For $$x=-3:$$ $$\beta=y=\dfrac{-3}{(-3)^{2}-3}=\dfrac{-3}{9-3}=\dfrac{-3}{6}=-\dfrac12.$$

Now we evaluate the combinations appearing in the options.

First try $$6\alpha+2\beta:$$

For $$(\alpha,\beta)=(3,\tfrac12):$$ $$6\alpha+2\beta=6\cdot3+2\cdot\dfrac12=18+1=19.$$

For $$(\alpha,\beta)=(-3,-\tfrac12):$$ $$6\alpha+2\beta=6\cdot(-3)+2\cdot\left(-\dfrac12\right)=-18-1=-19.$$

Among the given options, the numerical value $$19$$ matches exactly when $$\alpha=3,\;\beta=\dfrac12,$$ and it is presented in Option C:

$$6\alpha+2\beta=19.$$

Hence, the correct answer is Option C.