if A=(cosa,sina,0),B=(cosb,sinb,0),C=(cosc,sinc,0) are vertices of triangle abc and cosa+cosb+cosc=3a,sina+sinb+sinc=3b,then orthocentre is1.(3a,3b,0)2.(2a,2b,0)3.(0,0,0)4.(a,b,0)

dear student, this is a very ez ques. note that distance of all points A, B and C from the origin (0,0,0) is 1 bcoz sin^2theta+cos^2theta= 1. this means that A, B, C lie on the unit circle x^2+y^2=1, z= 0. Since the centre of this circle is (0,0,0), we conclude that the circumcentre of tri ABC is (0,0,0).

now, we know that euler line passes through orthocentre (H), centroid (G) and circumcentre (O) and divides G in the ratio 2:1. 

also, g= (cosa*i + sina*j + 0*k + cosb*i + sinb*j + 0*k + cosc*i + sinc*j + 0*k)/3

assume position vector of H as xi + yj + zk

ai + bj = (2(0i + 0j + 0k) + 1(xi + yj + zk))/3

or xi + yj + zk= 3ai + 3bj

so, x= 3a, y= 3b and z=0

so, H is (3a,3b,0)