$\begin{align}  & {{x}^{2}}=3\times 9=27\Rightarrow x=\sqrt{27}=3\sqrt{3} \\  & q=\frac{3\sqrt{3}}{3}=\sqrt{3} \\  & {{s}_{6}}=\frac{{{a}_{1}}(1-{{q}^{6}})}{1-q}=\frac{3(1-{{(\sqrt{3})}^{6}})}{1-\sqrt{3}}=\frac{3(1-27)}{1-\sqrt{3}} \\  & =\frac{3(-26)}{1-\sqrt{3}}\times \frac{1+\sqrt{3}}{1+\sqrt{3}}=\frac{3(-26)(1+\sqrt{3})}{1-3}=39(1+\sqrt{3}) \\ \end{align}$