If $3x+4y=12\sqrt{2}$ is a tangent to the ellipse  $\frac{x^2}{a^2}+\frac{y^2}{9}=1$ then the distance between its foci is

 

The equation of the tangent is

$3x+4y=12\sqrt{2}\Rightarrow y=\left(-\frac{3}{4}\right)x+3\sqrt{2}$

Using the condition of tendency $c^2=a^2{m}^2+b^2$ we obtain

$(3\sqrt{2}{)}^2=a^2{\left(-\frac{3}{4}\right)}^2+9\Rightarrow \frac{9}{16}{a}^2=9\Rightarrow a=4$

Let e be the eccentricity of the ellipse. Then

$e=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$

The distance between the foci is $2ae=2\times 4\times \frac{\sqrt{7}}{4}=2\sqrt{7}$