7. Given complex numbers $z_{1}, z_{2}, z_{3}$ satisfy:
$$
\begin{array}{l}
\left|z_{1}\right| \leqslant 1,\left|z_{2}\right| \leqslant 2, \\
\left|2 z_{3}-z_{1}-z_{2}\right| \leqslant\left|z_{1}-z_{2}\right| .
\end{array}
$$

Then the maximum value of $\left|z_{3}\right|$ is $\qquad$
7. $\sqrt{5}$.

Notice that,
$$
\begin{array}{l}
\left|2 z_{3}\right|-\left|z_{1}+z_{2}\right| \leqslant\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \\
\leqslant\left|z_{1}-z_{2}\right| .
\end{array}
$$

Then $2\left|z_{3}\right| \leqslant\left|z_{1}+z_{2}\right|+\left|z_{1}-z_{2}\right|$
$$
\begin{array}{l}
\leqslant \sqrt{2\left(\left|z_{1}+z_{2}\right|^{2}+\left|z_{1}-z_{2}\right|^{2}\right)} \\
=\sqrt{4\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)} \leqslant 2 \sqrt{5},
\end{array}
$$

where the equality holds when $z_{2}= \pm 2$ i $z_{1},\left|z_{1}\right|=1, z_{3}=z_{1}+z_{2}$. 
Therefore, the maximum value of $\left|z_{3}\right|$ is $\sqrt{5}$.