The equation $ a^2 b + b^2 c + c^2 a - ab^2 - bc^2 - ca^2 = 0 $ factors as \[(a - b)(b - c)(c - a) = 0.\]So, we want at least two of $ a, $ $ b, $ $ c $ to be equal. There are $ 100 \cdot 99 = 9900 $ triples $ (a,b,c) $ where $ a = b, $ and $ c $ is different from both $ a $ and $ b $. Similarly, there are 9900 triples where $ a = c, $ and $ b $ is different from both $ a $ and $ c, $ and 9900 triples where $ b = c, $ and $ a $ is different from both $ b $ and $ c, $ Finally, there are 100 triples of the form $ (a,a,a), $ so the total number of such triples is $ 3 \cdot 9900 + 100 = \boxed{29800} $.