11.166. The lengths of the edges of the parallelepiped are $a, b$, and $c$. The edges of lengths $a$ and $b$ are mutually perpendicular, and the edge of length $c$ forms an angle of $60^{\circ}$ with each of them. Determine the volume of the parallelepiped.
Solution.

In the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.60) $A B=a, A D=b$, $\angle B A D=90^{\circ}, A A_{1}=c, \angle A_{1} A B=\angle A_{1} A D=60^{\circ}, A_{1} K$ is the height. Since $\angle A_{1} A B=\angle A_{1} A D$, then $A K$ is the bisector of $\angle B A D . A A_{1}$ is the oblique to the base plane, $A K$ is its projection, $A D$ is a line lying in this plane. Then $\cos \angle A_{1} A D=\cos \angle A_{1} A K \cdot \cos \angle K A D$; $\cos \angle A_{1} A K=\frac{\cos \angle A_{1} A D}{\cos \angle K A D}=\frac{\cos 60^{\circ}}{\cos 45^{\circ}}=\frac{\sqrt{2}}{2}$.

Thus, $\angle A_{1} A K=45^{\circ}, \triangle A K A_{1}$ is a right and isosceles triangle and $A_{1} K=\frac{A A_{1} \sqrt{2}}{2}=\frac{c \sqrt{2}}{2}$. The volume of the parallelepiped $V=S_{A B C D} \cdot A_{1} K=\frac{a b c \sqrt{2}}{2}$.

Answer: $\frac{a b c \sqrt{2}}{2}$.