If O is the circumcentre, G is the centroid and O' is orthocentre or triangle ABC then prove that: `vec(OA) +vec(OB)+vec(OC)=vec(OO')`

Options

A. `vec(SA) + vec(SB) + vec(SC) =m 3vec(SG)`, where S is any point in the plane of `triangle ABC`.
B. `vec(OA) + vec(OB) +vec(OC) = vec(OO)`
C. `vec(O'A) + vec(O'B) + vec(O'C) = 2vec(O'O)`
D. `vec(AO') + vec(O'B) + vec(O'C) = vec(AP)`, where P is the diametrically opposite end to A.