6. $\frac{40 \sqrt{3}}{13}$.

According to the problem, we can construct a right triangle $\triangle ABC$ such that $|OA| = |z_1|$, $|OB| = |z_2|$, and $|OC| = |z_3|$. Then, $\angle ACB = 90^\circ$, $\angle AOC = 120^\circ$, $\angle BOC = 60^\circ$, $|BC| = 5$, and $|AC| = 12$. Thus, $|OA| + |OB| = 13$.
By $S_{\triangle ABC} = S_{\triangle BOC} + S_{\triangle AOC}$ $\Rightarrow \frac{1}{2} \times 5 \times 12$ $= \frac{\sqrt{3}}{4}(|OA||OC| + |OB||OC|)$ $\Rightarrow |OC|(|OA| + |OB|) = 40 \sqrt{3}$ $\Rightarrow |z_3| = |OC| = \frac{40 \sqrt{3}}{13}$.
6. In the complex plane, the complex numbers $z_{1}, z_{2}, z_{3}$ satisfy:
$$
\begin{array}{l}
z_{1}=\lambda z_{2}(\lambda<0),\left|z_{1}-z_{2}\right|=13, \\
\left|z_{1}\right|^{2}+\left|z_{3}\right|^{2}+\left|z_{1} z_{3}\right|=144, \\
\left|z_{2}\right|^{2}+\left|z_{3}\right|^{2}-\left|z_{2} z_{3}\right|=25 .
\end{array}
$$

Then $\left|z_{3}\right|=$