Given that,
$\log_{y}{x} = a\cdot \log_{z}{y} = b\cdot \log_{x}{z} = ab \longrightarrow (1)$
Let’s take first two terms.
$\log_{y}{x} = a\cdot \log_{z}{y}$
$\Rightarrow \boxed{a= \frac{\log_{y}{x}}{\log_{z}{y}}}$
And now, let’s take first and third terms.
$\log_{y}{x} = b \cdot \log_{x}{z}$
$\Rightarrow \boxed{b= \frac{\log_{y}{x}}{\log_{x}{z}}}$
Now, $ab = \left[\frac{\log_{y}{x}}{\log_{z}{y}}\right] \times \left[ \frac{\log_{y}{x}}{\log_{x}{z}} \right]$
$\Rightarrow ab = \left[\dfrac{\frac{\log_{c}{x}}{\log_{c}{y}}}{\frac{\log_{c}{y}}{\log_{c}{z}}}\right] \times \left[\dfrac{\frac{\log_{c}{x}}{\log_{c}{y}}}{\frac{\log_{c}{z}}{\log_{c}{x}}}\right] \quad \left[\because\log_{b}{a} = \frac{\log_{c}{a}}{\log_{c}{b}}\right]$
$\Rightarrow ab = \left[\frac{\log_{c}{x}}{\log_{c}{y}} \times \frac{\log_{c}{z}}{\log_{c}{y}}\right] \times \left[\frac{\log_{c}{x}}{\log_{c}{y}} \times \frac{\log_{c}{x}}{\log_{c}{z}}\right]$
$\Rightarrow ab = \left(\frac{\log_{c}{x}}{\log_{c}{y}}\right)^{3}$
$\Rightarrow ab = (\log_{y}{x})^{3}$
$\Rightarrow ab = (ab)^{3}$
$\Rightarrow ab – a^{3} b^{3} = 0$
$\Rightarrow ab (1- a^{2} b^{2}) = 0$
$ \Rightarrow \boxed{ab = 0} or 1- a^{2} b^{2} = 0$
$\Rightarrow a^{2} b^{2} = 1$
$\Rightarrow (ab)^{2} = 1$
$\Rightarrow \boxed{ab = \pm1}$
Now, we can check all of the options.
$A. (-2,\frac{1}{2}) \Rightarrow -2 \times \frac{1}{2} = -1$ (Possible)
$B. (1,1) \Rightarrow 1 \times 1 = 1$ (Possible)
$C. (\pi,\frac{1}{\pi}) \Rightarrow \pi \times \frac{1}{\pi} = 1$ (Possible)
$D. (2,2) \Rightarrow 2 \times 2 = 4$ (Not Possible)
Correct Answer $: \text{D}$