6 Given $\odot O_{1}\left(r_{1}\right), \odot O_{2}\left(r_{2}\right)$ are externally tangent to $\odot O(r)$ at $A_{1}, A_{2}$. When $r_{1}=1, r_{2}=2$, $r=3, A_{1} A_{2}=4$, find the length of the external common tangent $T_{1} T_{2}$ of $\odot O_{1}$ and $\odot O_{2}$.
\begin{array}{l}6 \cos \angle A_{1} O A_{2}=\frac{3^{2}+3^{2}-4^{2}}{2 \times 3 \times 3}=\frac{1}{9} \text{. } \\ O_{1} O_{2}^{2}=(3+1)^{2}+(3+2)^{2}-2(3+1)(3+2) \times \frac{1}{9} \\ =4^{2}+5^{2}-\frac{40}{9} \text{. } \\ T_{1} T_{2}=\sqrt{O_{1} O_{2}^{2}-(2-1)^{2}}=\sqrt{4^{2}+5^{2}-1-\frac{40}{9}} \\ =\frac{1}{3} \sqrt{40 \times 8}=\frac{8}{3} \sqrt{5} \text{. } \\\end{array}