(a) 1
\(\frac{1}{log_a\,bc+1}+\frac{1}{log_b\,ac+1}+\frac{1}{log_c\,ab+1}\)
= \(\frac{1}{log_a\,bc+log_a\,a}+\frac{1}{log_b\,ac+log_b\,b}+\frac{1}{log_c\,ab+log_C\,c}\)
= \(\frac{1}{log_a(abc)}+\frac{1}{log_b(abc)}+\frac{1}{log_c(abc)}\)
= logabc(a) + logabc(b) + logabc(c) = logabcabc = 1
\(\bigg(\)Using loga b = \(\frac{1}{log_ba}\bigg)\)