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

We are given the ellipse $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{9}=1$$ and the straight line $$3x+4y=12\sqrt{2}$$. The line is said to be a tangent to the ellipse, and we have to find the distance between the foci of that ellipse.

For any ellipse $$\dfrac{x^{2}}{A^{2}}+\dfrac{y^{2}}{B^{2}}=1$$ the condition that a straight line $$lx+my=n$$ should touch (be tangent to) the ellipse is the well-known tangent condition

$$n^{2}=A^{2}l^{2}+B^{2}m^{2}.$$

In our case we can match the coefficients as follows:

$$l=3,\; m=4,\; n=12\sqrt{2},$$

and, comparing $$(A^{2},B^{2})$$ with the given ellipse, we have

$$A^{2}=a^{2},\quad B^{2}=9.$$

Now we substitute these four numbers into the tangent condition:

$$\bigl(12\sqrt{2}\bigr)^{2}=a^{2}(3)^{2}+9(4)^{2}.$$

Simplifying each term step by step:

Left side:

$$\bigl(12\sqrt{2}\bigr)^{2}=12^{2}\times 2=144\times 2=288.$$

Right side:

$$a^{2}(3)^{2}+9(4)^{2}=a^{2}\cdot 9+9\cdot 16=9a^{2}+144.$$

Equating the two simplified sides we get

$$288=9a^{2}+144.$$

Now we isolate $$9a^{2}$$ by subtracting $$144$$ from both sides:

$$288-144=9a^{2},$$

so

$$144=9a^{2}.$$

Dividing both sides by $$9$$ yields

$$a^{2}=16.$$

Thus the semi-major axis squared of the ellipse is $$16$$. We observe that $$a^{2}=16$$ is greater than $$b^{2}=9$$, hence the major axis is along the $$x$$-direction. For an ellipse whose major axis is along the $$x$$-direction the focal distance relation is

$$c^{2}=a^{2}-b^{2}.$$

Substituting $$a^{2}=16$$ and $$b^{2}=9$$ gives

$$c^{2}=16-9=7.$$

Taking the positive square root,

$$c=\sqrt{7}.$$

The distance between the two foci is twice this value:

$$2c=2\sqrt{7}.$$

Hence, the correct answer is Option A.