Correct Answer: B To find the value of the given expression, we will simplify each term individually using fundamental properties of logarithms. First, let's analyze the term $1+x$: Given $x = \log_a (bc)$. We know that $1$ can be expressed as a logarithm with base $a$ as $\log_a a$. Therefore, $1+x = 1 + \log_a (bc) = \log_a a + \log_a (bc)$ Using the logarithm property $\log_k M + \log_k N = \log_k (MN)$, we combine the terms: $1+x = \log_a (a \cdot bc) = \log_a (abc)$ Now, we find the reciprocal $\frac{1}{1+x}$: $\frac{1}{1+x} = \frac{1}{\log_a (abc)}$ Applying the change of base property, which states that $\frac{1}{\log_M N} = \log_N M$: $\frac{1}{1+x} = \log_{abc} a$ (Equation 1) Next, let's analyze the term $1+y$: Given $y = \log_b (ca)$. Similarly, we express $1$ as $\log_b b$: $1+y = 1 + \log_b (ca) = \log_b b + \log_b (ca)$ Using the logarithm property $\log_k M + \log_k N = \log_k (MN)$: $1+y = \log_b (b \cdot ca) = \log_b (abc)$ Now, we find the reciprocal $\frac{1}{1+y}$: $\frac{1}{1+y} = \frac{1}{\log_b (abc)}$ Applying the change of base property: $\frac{1}{1+y} = \log_{abc} b$ (Equation 2) Finally, let's analyze the term $1+z$: Given $z = \log_c (ab)$. We express $1$ as $\log_c c$: $1+z = 1 + \log_c (ab) = \log_c c + \log_c (ab)$ Using the logarithm property $\log_k M + \log_k N = \log_k (MN)$: $1+z = \log_c (c \cdot ab) = \log_c (abc)$ Now, we find the reciprocal $\frac{1}{1+z}$: $\frac{1}{1+z} = \frac{1}{\log_c (abc)}$ Applying the change of base property: $\frac{1}{1+z} = \log_{abc} c$ (Equation 3) Now, we sum the three simplified terms: $\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z} = \log_{abc} a + \log_{abc} b + \log_{abc} c$ Using the logarithm property $\log_k M + \log_k N + \log_k P = \log_k (MNP)$: $= \log_{abc} (a \cdot b \cdot c)$ $= \log_{abc} (abc)$ Since $\log_K K = 1$ for any valid base $K$ ($K>0, K \neq 1$): $= 1$ Thus, the value of the entire expression is $1$. The final answer is $\boxed{1}$.