`A , B , Ca n dD` are any four points in the space, then prove that `| vec A Bxx vec C D+ vec B Cxx vec A D+ vec C Axx vec B D|=4` (area of ` A B C` .)

Explanation

Let P.V of A,B, C d and D be `veca, vecb,vecc and vec0` , respectively. Then `vec(AB)xxvec(CD)=(vecb-veca)xx(-vecc), vec(BC)xxvec(AD)=(vecc-vecb)xx(-veca)`
`vec(CA)xxvec(BD)=(veca-vecc)xx(-vecb)`
`vec(AB)xxvec(CD)+vec(BC)xxvec(AD)+vec(CA)xxvec(BD)`
`veccxxvecb+vecaxxvecc+vecaxxvecc+vecbxxveca-vecaxxvecb+veccxxvecb`
`2(veccxxvecb+ vecbxxvecbxxveca+vecaxxvecc)`
`2(veccxx(vecb-veca)-vecaxx(vecb-veca))`
`= 1(vec(AC)xxvec(AB))`
`|vec(AB)xxvec(CD)+vec(BC)xxvec(AD)+vec(CA)xxvec(BD)|=4|1/2(vec(AC)xxvec(AB))|=4triangleABC`