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The value of $\left|\begin{array}{ccc}1 & \log _{\mathrm{x}} \mathrm{y} & \log _{\mathrm{x}} \mathrm{z} \\ \log _{\mathrm{y}} \mathrm{x} & 1 & \log _{\mathrm{y}} \mathrm{z} \\ \log _{\mathrm{z}} \mathrm{x} & \log _{\mathrm{z}} \mathrm{y} & 1\end{array}\right|$ is equal to
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$\left|\begin{array}{ccc}1 & \log _{\mathrm{x}} \mathrm{y} & \log _{\mathrm{x}} \mathrm{z} \\ \log _{\mathrm{y} x} & 1 & \log _{\mathrm{y} z} \\ \log _{\mathrm{z}} \mathrm{x} & \log _{\mathrm{z}} \mathrm{y} & 1\end{array}\right|$
$=1\left(1-\log _{\mathrm{y}} \mathrm{z} \log _{\mathrm{z}} \mathrm{y}\right)$
$-\log _{\mathrm{x}} \mathrm{y}\left(\log _{\mathrm{y}} \mathrm{x}-\log _{\mathrm{z}} \mathrm{x} \log _{\mathrm{y}} \mathrm{z}\right)$
$\quad+\log _{\mathrm{x}} \mathrm{z}\left(\log _{\mathrm{z}} \mathrm{y}\left(\log _{\mathrm{y}} \mathrm{x}-\log _{\mathrm{z}} \mathrm{x}\right)\right.$
$=\left(1-\log _{\mathrm{y}} \mathrm{y}\right)-\log _{\mathrm{x} y}\left(\log _{\mathrm{y}} \mathrm{x}-\log _{\mathrm{y}} \mathrm{x}\right)$
$+\log _{\mathrm{x}} \mathrm{z}\left(\log _{\mathrm{z}} \mathrm{x}-\log _{\mathrm{z}} \mathrm{x}\right)$
$=(1-1)-0+0=0$
$=1\left(1-\log _{\mathrm{y}} \mathrm{z} \log _{\mathrm{z}} \mathrm{y}\right)$
$-\log _{\mathrm{x}} \mathrm{y}\left(\log _{\mathrm{y}} \mathrm{x}-\log _{\mathrm{z}} \mathrm{x} \log _{\mathrm{y}} \mathrm{z}\right)$
$\quad+\log _{\mathrm{x}} \mathrm{z}\left(\log _{\mathrm{z}} \mathrm{y}\left(\log _{\mathrm{y}} \mathrm{x}-\log _{\mathrm{z}} \mathrm{x}\right)\right.$
$=\left(1-\log _{\mathrm{y}} \mathrm{y}\right)-\log _{\mathrm{x} y}\left(\log _{\mathrm{y}} \mathrm{x}-\log _{\mathrm{y}} \mathrm{x}\right)$
$+\log _{\mathrm{x}} \mathrm{z}\left(\log _{\mathrm{z}} \mathrm{x}-\log _{\mathrm{z}} \mathrm{x}\right)$
$=(1-1)-0+0=0$
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