6. If the three sides of $\triangle A_{1} A_{2} A_{3}$ are $8, 10, 12$ respectively, and the midpoints of the three sides are $B, C, D$, connecting the midpoints in pairs to form three midlines, the resulting figure is the surface development of the tetrahedron $A-BCD$. Then, the surface area of the circumscribed sphere of this tetrahedron is
Answer $\frac{77 \pi}{2}$.
Analysis Given that the lengths of the three pairs of opposite edges of the tetrahedron $A-BCD$ are $4, 5, 6$.
Construct a rectangular prism such that the lengths of its face diagonals are $4, 5, 6$. Let the length, width, and height of the rectangular prism be $x, y, z$, and the radius of the circumscribed sphere be $R$, then
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=4^{2} \\
x^{2}+z^{2}=5^{2} \\
y^{2}+z^{2}=6^{2}
\end{array}\right.
$$

we get
$$
(2 R)^{2}=x^{2}+y^{2}+z^{2}=\frac{77}{2}
$$

Thus, $R^{2}=\frac{77}{8}$, so $S=\frac{77 \pi}{2}$.