If A`(veca).B(vecb) and C(vecc)` are three non-collinear point and origin does not lie in the plane of the points A, B and C, then for any point `P(vecP)` in the plane of the `triangleABC` such that vector `vec(OP)` is `bot ` to plane of `trianglABC`, show that `vec(OP)=([vecavecbvecc] (vecaxxvecb+vecbxxvecc+veccxxveca))/(4Delta^(2))`

Explanation

P lies in the plane of A,B and C , therefore,
`vec(AP).(vec(Bp) xx vec(CP))=0`
`Rightarrow (vecp.veca).(veccxxvecp+vecpxxvecb+vecbxxvecc)=0`
`or 0 +0+vecp . (vecbxxvecc)-veca.(veccxxvecp)`
`-veca.(vecpxxvecb)-veca.(vecbxxvecc)=0`
`or vecp . (vecbxxvecc)=vecp.(veccxxveca) +vecp.(vecaxxvecb)`
`-veca . (vecb xx vecc) =0`
`or vecp.(vecbxxvecc+veccxxveca+vecaxxvecb) = [veca vecbvecc]`
now vector perependicular to the plane ABC is
`(vecb xx vecc +vecc xx veca +veca xxvecb)`
Let ` vec(OP) = lambda(veca xx vecb + vecb +vecb xx vecc +vecc xx veca)`
since `vec(OP).(vecaxxvecb+vecbxxvecc+veccxxveca)=[veca vecbvecc]`
`or ([veca vecb vecc])/(4Delta^(2))`
`vec(OP) = ([veca vecb vecc] (veca xxvecb + vecb xxvecc +veccxxveca))/(4Delta^(2))`