Question:

Step 1: Convert logarithms using change of base formula. We'll express all logarithms in terms of natural logarithms using the change of base formula: $\log_b a = \frac{\ln a}{\ln b}$. Applying this to each term in the given equation: $$\log_b a \cdot \log_c a + \log_a b \cdot \log_c b + \log_a c \cdot \log_b c = \frac{\ln a}{\ln b} \cdot \frac{\ln a}{\ln c} + \frac{\ln b}{\ln a} \cdot \frac{\ln b}{\ln c} + \frac{\ln c}{\ln a} \cdot \frac{\ln c}{\ln b}$$ Step 2: Simplify the expression by introducing substitutions. Let $p = \ln a$, $q = \ln b$, and $r = \ln c$. The equation becomes: $$\frac{p^2}{qr} + \frac{q^2}{pr} + \frac{r^2}{pq} = 3$$ Step 3: Combine fractions over a common denominator. Multiply through by $pqr$ to get: $$\frac{p^3 + q^3 + r^3}{pqr} = 3$$ Therefore: $$p^3 + q^3 + r^3 = 3pqr$$ Step 4: Apply the algebraic identity for sum of cubes. We use the identity: $$p^3 + q^3 + r^3 - 3pqr = (p + q + r)(p^2 + q^2 + r^2 - pq - qr - rp)$$ Since $p^3 + q^3 + r^3 = 3pqr$, we have: $$p^3 + q^3 + r^3 - 3pqr = 0$$ This means: $$(p + q + r)(p^2 + q^2 + r^2 - pq - qr - rp) = 0$$ Step 5: Determine which factor equals zero. Since $a$, $b$, and $c$ are distinct positive real numbers not equal to 1, the values $p = \ln a$, $q = \ln b$, and $r = \ln c$ are distinct and non-zero. The expression $p^2 + q^2 + r^2 - pq - qr - rp = \frac{1}{2}[(p-q)^2 + (q-r)^2 + (r-p)^2]$ is always positive when $p$, $q$, $r$ are distinct. Therefore, we must have: $$p + q + r = 0$$ Step 6: Find the value of $abc$. From $p + q + r = 0$: $$\ln a + \ln b + \ln c = 0$$ $$\ln(abc) = 0$$ $$abc = e^0 = 1$$ **Final Answer:** $abc = 1$, which corresponds to **Option 2**.