\(2^{a}=3^{b}=6^{-c}=k\) (say) \[\] \(\Rightarrow 2=k^{\frac{1}{a}}, 3=k^{\frac{1}{b}}\) and \(k^{-\frac{1}{c}}\) \[\] As \(2 \times 3=6\) or \(k^{\frac{1}{a}} \cdot k^{\frac{1}{b}}=k^{-\frac{1}{c}}\) \[\] or \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\) \[\] \(\Rightarrow b c+a c=-a b\) or \(a b+b c+c a=0\) \[\] Choice (c)\[\]