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 $\left|z_{3}\right|=$ $\qquad$
6. $\frac{40 \sqrt{3}}{13}$.

From the problem, we can construct a right triangle $\triangle ABC$ such that
$$
|OA| = |z_1|, |OB| = |z_2|, |OC| = |z_3|.
$$

Then, $\angle ACB = 90^\circ$, $\angle AOC = 120^\circ$,
$$
\angle BOC = 60^\circ, |BC| = 5, |AC| = 12.
$$

Thus, $|OA| + |OB| = 13$.
By $S_{\triangle ABC} = S_{\triangle BOC} + S_{\triangle AOC}$,
$$
\begin{array}{l}
\Rightarrow \frac{1}{2} \times 5 \times 12 \\
= \frac{\sqrt{3}}{4} (|OA||OC| + |OB||OC|)
\end{array}
$$
$$
\begin{array}{l}
\Rightarrow |OC|(|OA| + |OB|) = 40 \sqrt{3} \\
\Rightarrow |z_3| = |OC| = \frac{40 \sqrt{3}}{13}
\end{array}
$$