OABC is a tetrahedron where O is the origin and A,B,C have position vectors `veca,vecb,vecc` respectively prove that circumcentre of tetrahedron OABC is `(a^2(vecbxxvecc)+b^2(veccxxveca)+c^2(vecaxxvecb))/(2[veca vecb vecc])`

Explanation

if the centre P is with position vector `vecr`, then
`veca-vecr = vec(PA) , vecb -vecr = vec(PB) , vecc -vecr= vec(PC)`,
`where|vec(PA)| = |vec(PB)|= |vec(PC)|= |vec(OP)|= |vecr|`
consider `|veca -vecr| = |vecr|`
`or (veca -vecr) . (veca -vecr) =vecr. vecr`
`or a^(2) - 2veca. vecr + r^(2) = r^(2) or a^(2) = 2veca. vecr`
similarly, `b^(2) = 2vecb. vecr, c^(2) = 2vecc. vecr`
since `(vecb xx vecc) , (vecc xx veca) + y (vecc xx vecc) + y.0 + z.0`
` x [veca vecb vecc]`
`or (veca.vecr)/([veca vecb vecc]) = a^(2) / (2[veca vecb vecc]) `
similarly, `y = b^(2)/(2[veca vecb vecc]) and z = c^(2)/(2[veca vecb vecc])`
Hence, `vecr= (a^(2) (vecb xxvecc) + vecb^(2) (veccxxveca) +c^(2) (veca xx vecb) )/(2[vecavecbvecc])`