11. (20 points) $z_{1}, z_{2}, z_{3}$ are the roots of the polynomial
$$
P(z)=z^{3}+a z+b
$$

satisfying
$$
\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+\left|z_{3}\right|^{2}=250,
$$

and the three points $z_{1}, z_{2}, z_{3}$ on the complex plane form a right triangle. Find the length of the hypotenuse of this right triangle.
11. By Vieta's formulas, we have
$$
z_{1}+z_{2}+z_{3}=0 \Rightarrow \frac{z_{1}+z_{2}+z_{3}}{3}=0
$$
$\Rightarrow$ The centroid of the triangle with vertices $z_{1}, z_{2}, z_{3}$ is the origin.

Without loss of generality, let $\left|z_{1}-z_{2}\right|=x,\left|z_{1}-z_{3}\right|=y$ be the two

legs of a right triangle. Since the distance from a vertex to the centroid is equal to $\frac{2}{3}$ of the length of the median from that vertex, we have
$$
\left|z_{1}\right|^{2}=\frac{4}{9}\left(x^{2}+\frac{y^{2}}{4}\right)=\frac{4}{9} x^{2}+\frac{1}{9} y^{2} .
$$

Similarly,
$$
\begin{array}{l}
\left|z_{2}\right|^{2}=\frac{4}{9}\left(y^{2}+\frac{x^{2}}{4}\right)=\frac{1}{9} x^{2}+\frac{4}{9} y^{2} . \\
\left|z_{3}\right|^{2}=\frac{4}{9}\left(\frac{x^{2}}{4}+\frac{y^{2}}{4}\right)=\frac{1}{9} x^{2}+\frac{1}{9} y^{2} . \\
\text { Then }\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+\left|z_{3}\right|^{2} \\
=\frac{6}{9} x^{2}+\frac{6}{9} y^{2}=\frac{2}{3} x^{2}+\frac{2}{3} y^{2}=250 \\
\Rightarrow \sqrt{x^{2}+y^{2}}=5 \sqrt{15} .
\end{array}
$$