11. If $\cot \alpha+\cot \beta+\cot \gamma=-\frac{4}{5}, \tan \alpha+\tan \beta+\tan \gamma=\frac{17}{6}$ and $\cot \alpha \cot \beta+\cot \beta \cot \gamma+\cot \gamma \cot \alpha=-\frac{17}{5}$, find the value of $\tan (\alpha+\beta+\gamma)$.
11. Answer: 11
Let $x=\tan \alpha, y=\tan \beta$ and $z=\tan \gamma$. Then
$$
\begin{aligned}
\frac{x y+y z+z x}{x y z} & =-\frac{4}{5} \\
x+y+z & =\frac{17}{6} \\
\frac{x+y+z}{x y z} & =-\frac{17}{5}
\end{aligned}
$$

From the above three equations, $x y z=-\frac{5}{6}$ and $x y+y z+z x=\frac{2}{3}$. It follows that
$$
\tan (\alpha+\beta+\gamma)=\frac{x+y+z-x y z}{1-(x y+y z+z x)}=\frac{17 / 6-(-5 / 6)}{1-2 / 3}=11 .
$$