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

$3x+4y=12\sqrt{2}\text{ is a tangent to the ellipse }\frac{x^2}{a^2}+\frac{y^2}{9}=1\Rightarrow a^2\left(3^2\right)+9\left(4^2\right)=(12\sqrt{2}{)}^2$

$\Rightarrow 9a^2+144=288\Rightarrow 9a^2=144\Rightarrow a^2=16$.

Now $e=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{16-9}}{4}=\frac{\sqrt{7}}{4}$. 

Distance between the foci  $=2ae=2\left(4\right)\frac{\sqrt{7}}{4}=2\sqrt{7}$.