Question
If $x = {\log _b}a,\,\,y = {\log _c}b,\,\,\,z = {\log _a}c$, then $xyz$ is
$ = {{{{\log }_e}a} \over {{{\log }_e}b}} \times {{{{\log }_e}b} \over {{{\log }_e}c}} \times {{{{\log }_e}c} \over {{{\log }_e}a}} = 1$.
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$\left|\begin{array}{ccc}
2 y+4 & 5 y+7 & 8 y+a \\
3 y+5 & 6 y+8 & 9 y+b \\
4 y+6 & 7 y+9 & 10 y+c
\end{array}\right|$