Assume \(14^a = 36^b = 84^c = k\).
Taking the logarithm of both sides yields:
\(\Rightarrow a = \log_{14} k \Rightarrow \frac{1}{a} = \log_k 14\)
Similarly, \( \frac{1}{c} = \log_k 84 \) and \( b = \log_{36} k \).
We need to evaluate:
\[6b\left( \frac{1}{c} - \frac{1}{a} \right)\]Substitute the expressions:
\[6 \cdot \log_{36} k \cdot (\log_k 84 - \log_k 14)\]Using the logarithm property \( \log_b a = \frac{1}{\log_a b} \), we rewrite the expression as:
\[6 \cdot \frac{\log k}{\log 36} \cdot \left( \frac{\log 84}{\log k} - \frac{\log 14}{\log k} \right)\]Simplify the expression:
\[6 \cdot \frac{\log k}{\log 36} \cdot \frac{\log 84 - \log 14}{\log k}\]\[= 6 \cdot \frac{\log 84 - \log 14}{\log 36}\]
\[= 6 \cdot \frac{\log \left( \frac{84}{14} \right)}{\log 36} = 6 \cdot \frac{\log 6}{\log 36}\]
Since \( \log 36 = \log 6^2 = 2 \log 6 \), the expression becomes:
\[6 \cdot \frac{\log 6}{2 \log 6} = 6 \cdot \frac{1}{2} = 3\]Final Answer: \( \boxed{3} \)