In $∆\mathrm{ABC} , \mathrm{sinA} \mathrm{sinB} \mathrm{sinC} = 1/3,$ then the value of $\mathrm{cotA} \mathrm{cotB} + \mathrm{cotB} \mathrm{cotC} + \mathrm{cotC} \mathrm{cotA} =$

$\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow A+B={180}^{\circ }-C\\ ⟹\cot (A+B)=\cot \left({180}^{\circ }-C\right)\\ \Rightarrow \frac{\cot A\cdot \cot B-1}{\cot A+\cot B}=-\cot C\\ \Rightarrow \cot A\cdot \cot B-1=-\cot A\cdot \cot C-\cot B\cdot \cot C\\ \Rightarrow \cot A\cdot \cot B+\cot B\cdot \cot C+\cot C\cdot \cot A=1\end{array}$