Example 7 Given $\tan \alpha+\tan \beta+\tan \gamma=\frac{17}{6}$, $\cot \alpha+\cot \beta+\cot \gamma=-\frac{4}{5}$, $\cot \alpha \cdot \cot \beta+\cot \beta \cdot \cot \gamma+\cot \gamma \cdot \cot \alpha=-\frac{17}{5}$.
Then $\tan (\alpha+\beta+\gamma)=$ $\qquad$ [5]
(2012, Xin Zhi Cup Shanghai High School Mathematics Competition)
Let $\tan \alpha=x, \tan \beta=y, \tan \gamma=z$.
Then $x+y+z=\frac{17}{6}$,
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{4}{5}$,
$\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=-\frac{17}{5}$.
(1) $\div$ (3) gives
$x y z=-\frac{5}{6}$.
(2) $\times$ (4) gives $x y+y z+z x=\frac{2}{3}$.

Therefore, $\tan (\alpha+\beta+\gamma)=\frac{x+y+z-x y z}{1-(x y+y z+z x)}=11$.