If \(x=\log _{a} b c, y=\log _{b} c a\) and \(z=\log _{c} a b\), then \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=\)
\(1+x=\log _{a} a+\log _{a} b c=\log _{a} a b c\) \[\] \(1+y=\log _{b} a b c\) \[\] \(1+z=\log _{c} a b c\), then \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\) \[\] \(=\log _{a b c} a+\log _{a b c} b+\log _{a b c} c=1\). \[\] Choice (C)\[\]
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