For the variable triangle ABC with fixed vertex (1,2) and A,B having coordinates (cost,sint),(sint,-cost) respectively, Find the locus of its centroid.

The coordinate of centroid O (x ,y ) in triangle having A (a,d ) , B(b,e) , C ( c ,f)
x = (a +b +c)/3 , y = (d + e + f)/3

So coordinate of given triangle is  :

h = (cost +sint +1)/3 , k = (sint - cost + 2 )/3

So 3h - 1 = cost + sint   (1) 
And 3k -2 =  sint - cost (2)

So squaring and adding both we have
(3h - 1)² + (3k -2)²  =  (cost + sint)² + (sint - cost)²
9h² +1 -6h + 9k² + 4 -12k = 2
9h² +3 -6h + 9k²  -12k = 0 
Replacing h and k by x and y , we have the locus eqaution
9x² +3 -6x + 9y²  -12y = 0