Three, (This question is worth 20 points) In the pyramid $A-B C D$, the three dihedral angles at vertex $D$ are all right angles, and the sum of the three dihedral angles at vertex $A$ is $90^{\circ}$. If $D B=a$, $D C=b$, find the volume of the pyramid $A-B C D$.
Three, let $A D=x$, unfold $A D, D B, D C$ on a plane, we get
$$
\begin{array}{l}
A D_{1}=A D_{2}=A D=x, \angle D_{1} A D_{2}=90^{\circ}, \\
\angle A D_{1} B=\angle A D_{2} C=90^{\circ}, \\
D_{1} B=a, D_{2} C=b, B C=\sqrt{a^{2}+b^{2}} .
\end{array}
$$

Extend $D_{1} B, D_{2} C$ to intersect at $E$, then $A D_{1} E D_{2}$ is a square with side length $x$. $D E=-a, C E=x-b$. By applying the Pythagorean theorem to $\mathrm{Rt} \triangle B E C$, we get
$$
a^{2}+b^{2}=(x-a)^{2}+(x-b)^{2} .
$$

Solving for $x$ yields $x=a+b$.
Therefore, the volume of the tetrahedron $V_{A} B C D=\frac{1}{6} a b(a+b)$.