II. (Full marks 30 points) In the tetrahedron $A-BCD$, $\angle DAB + \angle BAC + \angle DAC = 90^{\circ}$, $\angle ADB = \angle BDC = \angle ADC = 90^{\circ}$.
It is also known that $DB = a$, $DC = b$.
Try to find the radius of the largest sphere that can fit inside the tetrahedron $A-BCD$.
In the tetrahedron with a volume of 6, the sphere inscribed is its inner sphere.

Given $D B:=a, D C=6$, we set $D A=x$.
Cutting $D A, D B, D C$ and laying them out on a plane, $\angle D_{1} A D_{2}=90^{\circ}$, $D_{1} A=D_{2} A=x$. Extending $D_{1} B, D_{2} C$ to intersect at $K$, it is easy to see that $A D_{1} K D_{2}$ is a square. $B K=x-a, C K$
$=x-b, B C=$ $\sqrt{a^{2}+b^{2}}$.
By the Pythagorean theorem,
$$
B C^{2}=B K^{2}+
$$
$C K^{2}$,
which means $a^{2}+b^{2}=(x-a)^{2}+(x-b)^{2}$,
or $2 x^{2}=2(a+b) x$.
$$
\because x \neq 0, \quad \therefore x=a+b \text {. }
$$

Let the tetrahedron $A-B C D$ have an inscribed sphere with center $O$ and radius $r$.
Then $V_{A-B C D}=V_{O-A B D}+V_{O-A C D}+V_{O-A B C}+V_{O-B C D}$,
which means $\frac{1}{6} a b(a+b)=\frac{1}{3} \cdot r \cdot(a+b)^{2}$.
$$
\therefore \quad r=\frac{a b}{2(a+b)} .
$$