7. In the complex plane, the points corresponding to the complex numbers $z_{1}, z_{2}, z_{3}$ are $Z_{1}, Z_{2}, Z_{3}$, respectively. If
$$
\begin{array}{l}
\left|z_{1}\right|=\left|z_{2}\right|=\sqrt{2}, \overrightarrow{O Z_{1}} \cdot \overrightarrow{O Z_{2}}=0, \\
\left|z_{1}+z_{2}-z_{3}\right|=1,
\end{array}
$$

then the range of $\left|z_{3}\right|$ is $\qquad$
7. $[1,3]$.

Let $z_{1}=x_{1}+y_{1} \mathrm{i}, z_{2}=x_{2}+y_{2} \mathrm{i} (\mathrm{i}$ be the imaginary unit). Then
$$
\begin{array}{l}
x_{1}^{2}+y_{1}^{2}=x_{2}^{2}+y_{2}^{2}=2, x_{1} x_{2}+y_{1} y_{2}=0, \\
\left|z_{1}+z_{2}\right|=\sqrt{\left(x_{1}+y_{1}\right)^{2}+\left(x_{2}+y_{2}\right)^{2}} \\
=\sqrt{x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+2\left(x_{1} x_{2}+y_{1} y_{2}\right)}=2 .
\end{array}
$$

Let the point corresponding to the complex number $z_{1}+z_{2}$ be $P$.
From $\left|z_{1}+z_{2}-z_{3}\right|=1$, we know that the point $Z_{3}$ lies on the circle with center $P$ and radius 1.
Since $|O P|=2$, therefore, $2-1 \leqslant\left|O Z_{3}\right| \leqslant 2+1$, i.e., the range of $\left|z_{3}\right|$ is $[1,3]$.