7. Given complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=2,\left|z_{2}\right|=3$. If the angle between the vectors they correspond to is $60^{\circ}$, then $\left|\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right|=$ $\qquad$ .
By the cosine rule, we have $\left|z_{1}-z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-2\left|z_{1}\right| \cdot\left|z_{2}\right| \cos 60^{\circ}=4+9-6=7$, and $\left|z_{1}+z_{2}\right|^{2}+\left|z_{1}-z_{2}\right|^{2}=2\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)=26 \Rightarrow\left|z_{1}+z_{2}\right|^{2}=19$. Therefore, $\left|\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right|=\frac{\left|z_{1}+z_{2}\right|}{\left|z_{1}-z_{2}\right|}=\sqrt{\frac{19}{7}}=\frac{\sqrt{133}}{7}$.