Given: \[ 14^a = 36^b = 84^c = k \] Taking logarithms, we get: \[ a = \log_{14} k \implies \frac{1}{a} = \log_k 14 \] Similarly, \( b = \log_{36} k \) and \( \frac{1}{c} = \log_k 84 \).
The expression to evaluate is: \[ 6b \left( \frac{1}{c} - \frac{1}{a} \right) \]
Substitute the logarithmic forms: \[ 6 \cdot \log_{36} k \cdot (\log_k 84 - \log_k 14) \]
Apply the logarithm subtraction rule: \[ = 6 \cdot \log_{36} k \cdot \log_k \left( \frac{84}{14} \right) = 6 \cdot \log_{36} k \cdot \log_k 6 \]
Using the change of base formula, \( \log_{36} k = \frac{1}{\log_k 36} \). Substituting this gives: \[ = 6 \cdot \frac{1}{\log_k 36} \cdot \log_k 6 = 6 \cdot \frac{\log_k 6}{\log_k 36} \]
Note that \( \log_k 36 = \log_k (6^2) = 2 \log_k 6 \). Thus, the fraction simplifies to: \[ \frac{\log_k 6}{2 \log_k 6} = \frac{1}{2} \]
The final expression evaluates to: \[ 6 \cdot \frac{1}{2} = \boxed{3} \]
\(\boxed{3}\)