The locus of the point of intersection of the lines $$\left(\sqrt{3}\right)kx + ky - 4\sqrt{3} = 0$$ and $$\sqrt{3}x - y - 4\left(\sqrt{3}\right)k = 0$$ is a conic, whose eccentricity is ______

The two lines are $$\sqrt{3}\,kx + ky - 4\sqrt{3} = 0$$ and $$\sqrt{3}\,x - y - 4\sqrt{3}\,k = 0$$.

From the second equation, $$y = \sqrt{3}\,x - 4\sqrt{3}\,k$$, which gives $$k = \frac{\sqrt{3}\,x - y}{4\sqrt{3}}$$.

Substituting into the first equation: $$\sqrt{3}\left(\frac{\sqrt{3}\,x - y}{4\sqrt{3}}\right)x + \left(\frac{\sqrt{3}\,x - y}{4\sqrt{3}}\right)y = 4\sqrt{3}$$.

Simplifying the left side: $$\frac{x(\sqrt{3}\,x - y)}{4} + \frac{y(\sqrt{3}\,x - y)}{4\sqrt{3}} = 4\sqrt{3}$$.

Multiplying throughout by $$4\sqrt{3}$$: $$\sqrt{3}\,x(\sqrt{3}\,x - y) + y(\sqrt{3}\,x - y) = 48$$.

Expanding: $$3x^2 - \sqrt{3}\,xy + \sqrt{3}\,xy - y^2 = 48$$, which simplifies to $$3x^2 - y^2 = 48$$.

Dividing by 48: $$\frac{x^2}{16} - \frac{y^2}{48} = 1$$.

This is a hyperbola with $$a^2 = 16$$ and $$b^2 = 48$$. The eccentricity is $$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{48}{16}} = \sqrt{1 + 3} = \sqrt{4} = 2$$.

Therefore, the eccentricity of the conic is $$2$$.