This is a question from an old University of London UK A-Level question. I have solved most of the question but remain stuck on the last part (the angle/centroid) part (highlighted in red).

Here is the question:

The position vectors of the points A, B, C with respect to the origin O area, b, crespectively. If OA is perpendicular to BC, and OB is perpendicular to CA, show that OC is perpendicular to AB, and that OA^{2}+ BC^{2}= OB^{2}+ CA^{2}= OC^{2}+ AB^{2}.

Show that the plane through BC perpendicular to OA meets the plane through AB perpendicular to OC in a line that lies in the plane through OB perpendicular to CA.

If this line passes through the centroid of the triangle AOC, show that the angle AOC is pi/3 radians.

Here is my solution to the first (blue) parts of the question. Help with the final part greatly appreciated!

OA is perpendicular to BC, soa.(c - b)= 0

OB is perpendicular to CA, sob.(a - c)= 0

From these equations, we have

a.c-a.b= 0

b.a-b.c= 0

Adding these two expressions gives

a.c - a.b + b.a - b.c= 0

a.c - b.c= 0

soc.(a - b)= 0

Hence, OC is perpendicular to AB.

OA^{2}+ BC^{2}=a(^{2}+c-b)^{2}=a2^{2}+ b^{2}+ c^{2}-b.c

OB^{2}+ CA^{2}=b(^{2}+a-c)^{2}=a2^{2}+ b^{2}+ c^{2}-a.cOC

^{2}+ AB^{2}=c(^{2}+b-a)^{2}=a2^{2}+ b^{2}+ c^{2}-b.a

Butb.c = b.a, so OC^{2}+ AB^{2}= OA^{2}+ BC^{2}

anda.c = a.b, so OB^{2}+ CA^{2}= OC^{2}+ AB^{2}

Hence OA^{2}+ BC^{2}= OB^{2}+ CA^{2}= OC^{2}+ AB^{2}

The plane through BC perpendicular to OA has equation

r.a = b.a

and the plane through AB perpendicular to OC has equation

r.c = a.c

The line of intersection of these planes is perpendicular to bothaandc, i.e. it is parallel to CA. We know that OB is perpendicular to CA, so the equation of the line of intersection is given by

r=b+ t(c - a)

Then I grind to a halt...