ABC be a triangle with vertex A (0, 0), B(3, 1) and C (0, 2). If algebraic sum of perpendicular distance from A, B, C on a line mx+ny-1=0 is zero and this line passes through point (3, 5) then maximum value of k k∈N   such that 5m·3−n!   is divisible by 210k ?

Given that the algebraic sum of the perpendicular distances from A, B, C on a line mx+ny-1=0 is zero, this line is passing through the centroid of triangle ABC. It is also passing through the point (3,5)Hence the line mx +ny-1=0 is a line passing through the point G(1,1) and (3,5)Hence, m=2,n =-1752+7522+7523+...+7528=71,      therefore       10k=71 hence the nearest integer value of k is 7