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

$$\begin{gathered} The line\frac{\mathrm{x}}{7}+\frac{\mathrm{y}}{2\sqrt{6}}=1 intersects x- axis at \left(7,0\right) and \\ \frac{\mathrm{x}}{7}-\frac{\mathrm{y}}{2\sqrt{6}}=1 intersects y- axis at \left(0,-2\sqrt{6}\right). \\ substituting \left(7,0\right) and \left(0,-2\sqrt{6}\right). in \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \Rightarrow a=7,b=2\sqrt{6} \\ \therefore e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{24}{49}}=\frac{5}{7} \end{gathered}$$