Given that , $\log_{3}{x} = \log_{12}{y} = a \quad \longrightarrow (1) $
From equation $(1),\;\log_{3}{x} = a$
$\Rightarrow \boxed{x = 3^{a}} \quad [\because \log_{a}{x} = b \Rightarrow x = a^{b}]$
Again, from equation $(1),\;\log_{12}{y} = a$
$\Rightarrow \boxed{y = 12^{a}}$
If $G$ is the geometric mean of $x$ and $y$, then $G= \sqrt{xy}$
$\Rightarrow G = \sqrt{3^{a}\cdot 12^{a}} $
$\Rightarrow G = \sqrt{3^{a}\cdot (3 \cdot 4)^{a}}$
$\Rightarrow G= \sqrt{3^{a}\cdot 3^{a}\cdot4^{a}}$
$\Rightarrow G= \sqrt{3^{2a}\cdot 2^{2a}}$
$\Rightarrow G= 3^{a}\cdot 2^{a}$
$\Rightarrow \boxed{G= 6^{a}}$
$\therefore$ The value of $\log_{6}{G} = \log_{6}{6^{a}} = a \log_{6}{6} = a$
Correct Answer $ : \text{D}$
$\textbf{PS:}$
- $\log_{b}{a^{x}} = x\log_{b}{a}$
- $\log_{a}{a} = 1$