If $\frac{\tan (\ln 6)\cdot \tan (\ln 2)\cdot \tan (\ln 3)}{\tan (\ln 6)-\tan (\ln 2)-\tan (\ln 3)}=\mathrm{k}$, then the value of k is _______.

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

If $\mathrm{\gamma }=\mathrm{\alpha }+\mathrm{\beta }$, then

$\tan \mathrm{\gamma }-\tan \mathrm{\alpha }-\tan \mathrm{\beta }=\tan \mathrm{\gamma }\cdot \tan \mathrm{\alpha }\cdot \tan \mathrm{\beta }$

$\Rightarrow \frac{\tan (\ln 6)\cdot \tan (\ln 2)\cdot \tan (\ln 3)}{\tan (\ln 6)-\tan (\ln 2)-\tan (\ln 3)}=1$