Let $$l_1$$ be the line in $$xy$$-plane with $$x$$ and $$y$$ intercepts $$\frac{1}{8}$$ and $$\frac{1}{4\sqrt{2}}$$ respectively, and $$l_2$$ be the line in $$zx$$-plane with $$x$$ and $$z$$ intercepts $$-\frac{1}{8}$$ and $$-\frac{1}{6\sqrt{3}}$$ respectively. If $$d$$ is the shortest distance between the line $$l_1$$ and $$l_2$$, then $$d^{-2}$$ is equal to ______.

Line $$l_1$$ is in the $$xy$$-plane with $$x$$-intercept $$\frac{1}{8}$$ and $$y$$-intercept $$\frac{1}{4\sqrt{2}}$$, so it passes through $$A_1 = \left(\frac{1}{8}, 0, 0\right)$$ and $$B_1 = \left(0, \frac{1}{4\sqrt{2}}, 0\right)$$. Its direction vector is $$\vec{d_1} = B_1 - A_1 = \left(-\frac{1}{8}, \frac{1}{4\sqrt{2}}, 0\right)$$, and multiplying by $$-8$$ gives $$\vec{d_1} = (1, -\sqrt{2}, 0)$$.

Similarly, line $$l_2$$ in the $$zx$$-plane has $$x$$-intercept $$-\frac{1}{8}$$ and $$z$$-intercept $$-\frac{1}{6\sqrt{3}}$$, so it passes through $$A_2 = \left(-\frac{1}{8}, 0, 0\right)$$ and $$B_2 = \left(0, 0, -\frac{1}{6\sqrt{3}}\right)$$. Its direction vector is $$\vec{d_2} = B_2 - A_2 = \left(\frac{1}{8}, 0, -\frac{1}{6\sqrt{3}}\right)$$, and multiplying by $$24\sqrt{3}$$ yields $$\vec{d_2} = (3\sqrt{3}, 0, -4)$$.

Next, the cross product is computed as $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -\sqrt{2} & 0 \\ 3\sqrt{3} & 0 & -4 \end{vmatrix}$$, which gives
$$\hat{i}: (-\sqrt{2})(-4) - (0)(0) = 4\sqrt{2}$$
$$\hat{j}: -[(1)(-4) - (0)(3\sqrt{3})] = -(-4) = 4$$
$$\hat{k}: (1)(0) - (-\sqrt{2})(3\sqrt{3}) = 3\sqrt{6}$$, so $$\vec{d_1} \times \vec{d_2} = 4\sqrt{2}\,\hat{i} + 4\hat{j} + 3\sqrt{6}\,\hat{k}$$.

Then its squared magnitude is $$|\vec{d_1} \times \vec{d_2}|^2 = (4\sqrt{2})^2 + 4^2 + (3\sqrt{6})^2 = 32 + 16 + 54 = 102$$.

Moreover, the vector joining the points on the two lines is $$\vec{A_2 - A_1} = \left(-\frac{1}{8} - \frac{1}{8}, 0, 0\right) = \left(-\frac{1}{4}, 0, 0\right)$$. Since the distance between two skew lines is given by $$d = \frac{|(\vec{A_2 - A_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$$, one computes the numerator as $$(\vec{A_2 - A_1}) \cdot (\vec{d_1} \times \vec{d_2}) = \left(-\frac{1}{4}\right)(4\sqrt{2}) + (0)(4) + (0)(3\sqrt{6}) = -\sqrt{2}$$ so $$|(\vec{A_2 - A_1}) \cdot (\vec{d_1} \times \vec{d_2})| = \sqrt{2}$$. Therefore, $$d = \frac{\sqrt{2}}{\sqrt{102}} = \sqrt{\frac{2}{102}} = \sqrt{\frac{1}{51}} = \frac{1}{\sqrt{51}}$$.

Finally, since $$d^2 = \frac{1}{51}$$ one finds $$d^{-2} = 51$$, and thus the answer is $$\boxed{51}$$.