If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA² +PB²+ PC² = GA² + GB²+ GC² + 3GP².

You can do it by coordinates and vectors both methods

vector method is simple and involves less variables

assume P as origin

Position vectors of A,B,C as a,b,c respectively

PA=a, PB=b,OPC=c

G=a+b+c/3

GA=a-(a+b+c)/3=2a-b-c/3

similarly you can write for GB,GC,GP

GA2=(4a2+b2+c2-4a.b+b.c-4a.c)/9

similarly write for others and just calculate the expression

similarly using coordinate 

you can take P as (0,0)and A,B,C as (x1,y1),(x2,y2)and (x3,y3)

G will be ((x1+x2+x3)/3 , (y1+y2+y3)/3)

and using distance formula solve the expressions on LHS and RHS