jmc

The equation $a^{2}b+b^{2}c+c^{2}a-a{b}^2-b{c}^2-c{a}^2=0$ 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 $3\cdot 9900+100=\boxed{29800}.$