Answer: $\mathbf A$
Explanation:
We know that:
$\color {blue} {\mathbf {\log_ab = \frac{\log b}{\log a}}}$
$\therefore$ On solving the determinant we, get:
$ \mathrm {1(1-\log_zy\log_yz) -\log_xy(1.\log_yx-\log_zx\log_xz) + \log_xz(1.\log_zx-\log_yx\log_zy)}$
$= \mathrm {1\left (1-\frac{{\log y}}{\log z}\frac{\log z}{\log y}\right ) - \frac{\log y}{\log x}\left (\frac{\log x}{\log y}-\frac{\log x}{\log z}\frac{\log z}{\log x}\right ) + \frac{\log z}{\log x}\left ( \frac{\log x}{\log z}-\frac{\log x}{\log y}\frac{\log y}{\log z}\right) }$
$=0$
$\therefore \mathbf A$ is the correct option.