If $\frac{\tan \left(\ln 6\right).\tan \left(\ln 2\right).\tan \left(\ln 3\right)}{\tan \left(\ln 6\right)-\tan \left(\ln 2\right)-\tan \left(\ln 3\right)}=k,$  then the value of  $k=$__________

$\ln 6=\ln 3+\ln 2$

 If  $\gamma =\alpha +\beta ,$ then 
 $$\begin{gathered} \tan \gamma -\tan \alpha -\tan \beta =\tan \gamma .\tan \alpha .\tan \beta \\ \Rightarrow \frac{\tan \left(\ln 6\right).\tan \left(\ln 2\right).\tan \left(\ln 3\right)}{\tan \left(\ln 6\right)-\tan \left(\ln 2\right)-\tan \left(\ln 3\right)}=1 \end{gathered}$$