If the two diagonals of one of its faces are 6i^+6k^ and 4j^+2k^  and of the edges not containing the given diagonals c→=4j^−8k^ then the volume of a parallelepiped is

Let a→=6i^+6k^,b→=4j^+2k^,c→=4j^−8k^ Then a→×b→=−24i^−12j^+24k^                    =12(−2i^−j^+2k^)Area of the base of the parallelepiped =12|a→×b→|=12(12×3)=18Height of the parallelepiped = Length of projection of c→ on a→×b→=|c→⋅a→×b→||a→×b→|=|12(−4−16)|36=203∴ Volume of the parallelepiped =18×203=120