Complex numbers $a$, $b$ and $c$ are the zeros of a polynomial $P(z) = z^3+qz+r$, and $|a|^2+|b|^2+|c|^2=250$.  The points corresponding to $a$, $b$, and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h$.  Find $h^2$.
1. Given that \(a\), \(b\), and \(c\) are the zeros of the polynomial \(P(z) = z^3 + qz + r\), we know from Vieta's formulas that:
   \[
   a + b + c = 0
   \]
   \[
   ab + bc + ca = q
   \]
   \[
   abc = -r
   \]

2. We are also given that:
   \[
   |a|^2 + |b|^2 + |c|^2 = 250
   \]

3. Since \(a\), \(b\), and \(c\) form a right triangle, we can assume without loss of generality that \(a\) and \(b\) are the legs and \(c\) is the hypotenuse. This implies:
   \[
   |c|^2 = |a|^2 + |b|^2
   \]

4. Using the given condition \( |a|^2 + |b|^2 + |c|^2 = 250 \) and the fact that \( |c|^2 = |a|^2 + |b|^2 \), we can substitute:
   \[
   |a|^2 + |b|^2 + (|a|^2 + |b|^2) = 250
   \]
   \[
   2(|a|^2 + |b|^2) = 250
   \]
   \[
   |a|^2 + |b|^2 = 125
   \]

5. Since \(c\) is the hypotenuse, we have:
   \[
   |c|^2 = |a|^2 + |b|^2 = 125
   \]

6. Therefore, the square of the hypotenuse \(h\) is:
   \[
   h^2 = |c|^2 = 125
   \]

The final answer is \(\boxed{125}\)