7. Let $x, y, z$ be complex numbers, and
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=x y+y z+z x, \\
|x+y+z|=21,|x-y|=2 \sqrt{3},|x|=3 \sqrt{3} .
\end{array}
$$

Then $|y|^{2}+|z|^{2}=$ . $\qquad$
7. 132.

It is easy to see that the figures corresponding to $x$, $y$, and $z$ on the complex plane form an equilateral triangle, and note that
\[
\begin{array}{l}
|x-y|^{2}+|y-z|^{2}+|z-x|^{2}+|x+y+z|^{2} \\
=3\left(|x|^{2}+|y|^{2}+|z|^{2}\right),
\end{array}
\]

Combining the conditions and $|x-y|=|y-z|=|z-x|$, we can substitute and solve.