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

$$

\dfrac{x}{7} = 1 \implies x = 7

$$

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

$$

\dfrac{49}{a^2} = 1 \implies a^2 = 49, \quad a = 7

$$

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

$$

-\dfrac{y}{2\sqrt{6}} = 1 \implies y = -2\sqrt{6}

$$

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

$$

\dfrac{24}{b^2} = 1 \implies b^2 = 24

$$

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

$$

e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{24}{49}} = \sqrt{\dfrac{25}{49}} = \dfrac{5}{7}

$$

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