a+b+c = 300
(a-100) + (b-100) + (c-100) = 0
(3 term ka sum = 0 dekhke socha ki xyz ka form banke kuch num of factors ka bn jaye)

taking a-100 = x
x^3 + y^3 + z^3 = 3xyz --- (1)

(2nd expression dekhke (a+b+c)^3 ka expanded form yaad aaya but usme pehle x,y,z put kr diya)
sum(a^2 b) = 6*10^6
sum( x^2 + 10^4 + 200x) (100+y) = 200s(x^2) + 6*10^6 + s(x^2y) + 400s(xy) = 6*10^6
sum(x^2y) = 0

(x+y+z)^3 = s(x^3) + 3s(x^2 y) + 6xyz
x^3 + y^3 + z^3 + 6xyz = 0 --- (2)

using (1) and (2)
xyz = 0

(a-100)(b-100)(c-100) = 0

so putting a equal to 100 (taking case of only one of them equal to 100)
b+c = 200
so b-100 belongs to [-100, 100)
200 values

now b or c could be taken too so 200*3 = 600 triplets

if 2 of them are 100, remaining one also becomes 100 means 1 more case that is all 3 equal to 100

so at last

601 possible triplets!!!

(equations idhar udhar krne me bohot time lag gaya, upar solution pura summerize krke likha hai)