Correct Answer: B) $1$ To determine the value of the given expression, we will simplify each term individually. First, consider the term $1+x$: $$\begin{aligned} 1+x &= 1 + \log_a (bc) \\ &= \log_a a + \log_a (bc) \quad \text{(Since } 1 = \log_a a\text{)} \\ &= \log_a (a \cdot bc) \quad \text{(Using the product rule: } \log_M N + \log_M P = \log_M (NP)\text{)} \\ &= \log_a (abc) \end{aligned}$$ Now, let's find the reciprocal of $1+x$: $$\frac{1}{1+x} = \frac{1}{\log_a (abc)}$$ Using the change of base property of logarithms, which states that $\frac{1}{\log_M N} = \log_N M$, we can rewrite this as: $$\frac{1}{1+x} = \log_{abc} a$$ Similarly, for the term $1+y$: $$\begin{aligned} 1+y &= 1 + \log_b (ac) \\ &= \log_b b + \log_b (ac) \\ &= \log_b (b \cdot ac) \\ &= \log_b (abc) \end{aligned}$$ And its reciprocal: $$\frac{1}{1+y} = \frac{1}{\log_b (abc)} = \log_{abc} b$$ Finally, for the term $1+z$: $$\begin{aligned} 1+z &= 1 + \log_c (ab) \\ &= \log_c c + \log_c (ab) \\ &= \log_c (c \cdot ab) \\ &= \log_c (abc) \end{aligned}$$ And its reciprocal: $$\frac{1}{1+z} = \frac{1}{\log_c (abc)} = \log_{abc} c$$ Now, we sum these simplified terms: $$\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z} = \log_{abc} a + \log_{abc} b + \log_{abc} c$$ Applying the product rule for logarithms again (since the bases are now the same, $abc$): $$= \log_{abc} (a \cdot b \cdot c)$$ $$= \log_{abc} (abc)$$ Since $\log_M M = 1$ for any valid base $M$: $$= 1$$ Thus, the value of the expression is $1$.