Find the value of $\frac{\mathrm{z}(\mathrm{x}+2\mathrm{y})}{\mathrm{xy}}$ if $2^{\mathrm{x}}=3^{\mathrm{y}}={12}^{\mathrm{z}}$.

We have:
 $2^{\mathrm{x}}=3^{\mathrm{y}}={12}^{\mathrm{z}}=\mathrm{k}$
Then,
$2={\mathrm{k}}^{\frac{1}{\mathrm{x}}},\mathrm{ }3={\mathrm{k}}^{\frac{1}{\mathrm{y}}}$  and $12={\mathrm{k}}^{\frac{1}{\mathrm{z}}}$

We know that:

$12=2^2\times 3$

Therefore,
$\Rightarrow {\mathrm{k}}^{\frac{1}{\mathrm{z}}}={\mathrm{k}}^{\frac{2}{\mathrm{x}}}\times {\mathrm{k}}^{\frac{1}{\mathrm{y}}}$

$\Rightarrow {\mathrm{k}}^{\frac{1}{\mathrm{z}}}={\mathrm{k}}^{\frac{2\mathrm{y}+\mathrm{x}}{\mathrm{xy}}}$

$\Rightarrow \frac{1}{\mathrm{z}}=\frac{2\mathrm{y}+\mathrm{x}}{\mathrm{xy}}$
$\text{⇒}\frac{z\left(2y+x\right)}{xy}=1$

Hence, $\frac{\mathbf{z}\mathbf{(}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{x}\mathbf{)}}{\mathbf{xy}}\mathbf{=}\mathbf{1}$.