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 the of ellipse is

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

Condition of tangency $c^2=a^2{m}^2+b^2$

$\Rightarrow 18=a^2.\frac{9}{16}+9$

$\Rightarrow a^2.\frac{9}{16}=9$

$\Rightarrow a^2=16$

$\Rightarrow a=4$

$\Rightarrow e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$

$\therefore ae=\frac{\sqrt{7}}{4}.4=\sqrt{7}$

 $\Rightarrow$foci are   $\left(\pm \sqrt{7},0\right)$

$\therefore$ distance between foci $=2\sqrt{7}$