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By Vieta's formula, the sum of the roots is equal to 0, or $a+b+c=0$. Therefore, $\frac{a+b+c}{3}=0$. Since the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin.

Without loss of generality, let the right angle be at $b.$ Let $x=|b-c|$ and $y=|a-b|.$ The magnitudes of $a$, $b$, and $c$ are just $\frac{2}{3}$ of the medians because the origin, or the centroid in this case, cuts the median in a ratio of $2:1$.

Hence, $$|a{|}^2=\frac{4}{9}\cdot \left({\left(\frac{x}{2}\right)}^2+y^2\right)=\frac{x^2}{9}+\frac{4y^2}{9}$$because $|a|$ is two thirds of the median from $a$. Similarly, $$|c{|}^2=\frac{4}{9}\cdot \left(x^2+{\left(\frac{y}{2}\right)}^2\right)=\frac{4x^2}{9}+\frac{y^2}{9}.$$Furthermore, $$|b{|}^2=\frac{4}{9}\cdot \frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}.$$Hence, $$|a{|}^2+|b{|}^2+|c{|}^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250.$$Thus, $h^2=x^2+y^2=\frac{3}{2}\cdot 250=\boxed{375}.$