Given that, $a,b,c$ are non-zero.
And, $ 14^{a} = 36^{b} = 84^{c} $
We can do the prime factorization.
$ \Rightarrow (2 \times 7)^{a} = (4 \times 9)^{b} = (7 \times 12)^{c} $
$ \Rightarrow 2^{a} \times 7^{a} = (2^{2} \times 3^{2})^{b} = (7 \times 2^{2} \times 3)^{c}$
$ \Rightarrow 2^{a} \times 7^{a} = 2^{2b} \times 3^{2b} = 7^{c} \times 2^{2c} \times 3^{c} $
On comparing, we get, $ a = 2b = 2c $
Then, $ 6b \left( \frac{1}{c} – \frac{1}{a} \right) = 6b \left( \frac{1}{b} – \frac{1}{2b} \right) = 6b \left( \frac{2-1}{2b} \right) = 3 $
$\therefore$ The value of $ 6b \left( \frac{1}{c} – \frac{1}{a} \right) = 3.$
$\textbf{Second Method:}$
Let $ 14^{a} = 36^{b} = 84^{c} = x \; \longrightarrow (1) $
Now, $14^{a} = x$
Taking $\log_{14}$ on both sides
$ \log_{14} 14^{a} = \log_{14}x $
$ \Rightarrow a \log_{14} 14 = \log_{14}x \quad [ \because \log_{a}b^{x} = x \log_{a}b] $
$ \Rightarrow \boxed{a = \log_{14}x} \quad [ \because \log_{a}a = 1] $
$ \Rightarrow \boxed{\frac{1}{a} = \log_{x}14} \quad \left[ \because \log_{a}b = \frac{1}{\log_{b}a}\right] $
Again, $36^{b} = x$
Taking $\log_{36}$ on both sides.
$ \log_{36}36^{b} = \log_{36}x $
$ \Rightarrow \boxed{b = \log_{36}x} $
Again, $ 84^{c} = x $
Taking $\log_{84}$ on both sides.
$ \log_{84}84^{c} = \log_{84}x $
$ \Rightarrow \boxed{c = \log_{84}x} $
$ \Rightarrow \boxed{\frac{1}{c} = \log_{x}84} $
Then, $ 6b \left( \frac{1}{c} – \frac{1}{a} \right) = 6 \log_{36}x \left( \log_{x}84 – \log_{x}14 \right) $
$\qquad = 6 \log_{36}x \left[ \log_{x} \left( \frac{84}{14} \right) \right] \quad \left[ \because \log_{a}m – \log_{a}n = \log_{a} \left( \frac{m}{n} \right)\right] $
$\qquad = 6 \log_{36}x \cdot \log_{x}6 $
$\qquad = 6 \left( \frac{\log_{c}x}{\log_{c}36} \right) \left( \frac{\log_{c}6}{\log_{c}x} \right) \quad \left[ \because \log_{m}n = \frac{\log_{c}n}{\log_{c}m}\right] $
$\qquad = 6 \left( \frac{\log_{c}6}{\log_{c}6^{2}} \right) $
$\qquad = 6 \left( \frac{\log_{c}6}{2 \log_{c}6} \right) $
$\qquad = 3 $
Correct Answer$: 3$