If (2, -3), (3, 7) and (-4, 6) are the mid points of the sides BC, CA and AB respectively of the $$\triangle ABC$$, then A =

Let co-ordinate of A = (x1,y1), B = (x2, y2) and C = (x3, y3)

Mid-point of AB = $$\frac{x1 + x2}{2}$$,$$\frac{y1 + y2}{2}$$ = (-4, 6)

x1 + x2 = -8 and y1 + y2 = 12 ............. (1)

Similarly, 

x2 + x3 = 4 and y2 + y3 = -6 ............. (2)

x1 + x3 = 6 and y1 + y3  = 14 ............. (3)

From (1), (2) and (3):

x1 + x2 + x3 = 1 and y1 + y2 + y3 = 10 .......... (4)

From (2) and (4):

x1 = -3 and y1 = 16

Hence, A = (-3, 16)