If the ellipse $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ meets the line $$\dfrac{x}{7} + \dfrac{y}{2\sqrt{6}} = 1$$ on the $$x$$-axis and the line $$\dfrac{x}{7} - \dfrac{y}{2\sqrt{6}} = 1$$ on the $$y$$-axis, then the eccentricity of the ellipse is

We need to find the eccentricity of the ellipse $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$. First we determine where the line $$\dfrac{x}{7} + \dfrac{y}{2\sqrt{6}} = 1$$ meets the $$x$$-axis by setting $$y = 0$$, which reduces to

Therefore the ellipse passes through $$(7, 0)$$. Substituting these coordinates into the ellipse equation gives

Next, to find where the line $$\dfrac{x}{7} - \dfrac{y}{2\sqrt{6}} = 1$$ meets the $$y$$-axis, we set $$x = 0$$, yielding

This means the ellipse passes through $$(0, -2\sqrt{6})$$. Substituting this point into the ellipse equation gives

Since $$a^2 = 49 > b^2 = 24$$, the major axis lies along the $$x$$-axis. Hence the eccentricity is

The correct answer is Option A: $$\dfrac{5}{7}$$.