6. 8 .

As shown in Figure 3, construct the circumscribed parallelepiped $A A_{1} B B_{1}-D_{1} C C_{1} D$ of the tetrahedron $A B C D$. Since $E F \perp C D$, $C D / / A_{1} B_{1}$, then $E F \perp A_{1} B_{1}$. Also, $E F \perp A B$, so $E F \perp$ the plane $A A_{1} B B_{1}$. Since $E F / / A_{1} C / / B_{1} D$, it follows that $C A_{1} \perp$ the plane $A A_{1} B B_{1}$. Therefore, this parallelepiped is a right parallelepiped.
It is easy to see that $A A_{1}=8$. Also, $B C=6 \sqrt{2}$, so $A_{1} B=6$, and $\angle A A_{1} B=90^{\circ}$. Therefore, this parallelepiped is a rectangular parallelepiped. Hence, the distance between $A D$ and $B C$ is $A A_{1}=8$.
6. In tetrahedron $ABCD$, the opposite edges are equal, $E, F$ are the midpoints of $AB, CD$ respectively, and $EF \perp AB, EF \perp CD, EF=6, AC=10, BC=6\sqrt{2}$. Then the distance between the skew lines $AD$ and $BC$ is $\qquad$.