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(A) Let $a$ be the first term and $R$ be the common ratio of the geometric progression $(GP)$.Then,the $n$-th term is given by $T_n = a R^{n-1}$.Given

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(A) Let $a$ be the first term and $R$ be the common ratio of the geometric progression $(GP)$.Then,the $n$-th term is given by $T_n = a R^{n-1}$.Given that $x, y, z$ are the $p$-th,$q$-th,and $r$-th terms respectively:$x = a R^{p-1}$,$y = a R^{q-1}$,$z = a R^{r-1}$.Taking the logarithm on both sides:$\log x = \log a + (p-1) \log R$$\log y = \log a + (q-1) \log R$$\log z = \log a + (r-1) \log R$Now,substitute these into the determinant $\Delta = \left|\begin{array}{lll} \log x & p & 1 \ \log y & q & 1 \ \log z & r & 1 \end{array}ight|$.$\Delta = \left|\begin{array}{lll} \log a + (p-1) \log R & p & 1 \ \log a + (q-1) \log R & q & 1 \ \log a + (r-1) \log R & r & 1 \end{array}ight|$.Using the property of determinants,we can split this into two determinants:$\Delta = \left|\begin{array}{lll} \log a & p & 1 \ \log a & q & 1 \ \log a & r & 1 \end{array}ight| + \left|\begin{array}{lll} (p-1) \log R & p & 1 \ (q-1) \log R & q & 1 \ (r-1) \log R & r & 1 \end{array}ight|$.In the first determinant,the first column is $\log a$ times the third column,so it is $0$.In the second determinant,perform the operation $C_1 ightarrow C_1 - C_3 \log R$ (or simply observe that $C_1$ is proportional to $C_2 - C_3$):Specifically,$C_1 = (C_2 - C_3) \log R$. Since $C_1$ is a multiple of $(C_2 - C_3)$,the determinant is $0$.Thus,$\Delta = 0 + 0 = 0$.

If $x, y, z$ are all positive and are the $p$-th,$q$-th and $r$-th terms of a geometric progression respectively,then the value of the determinant $\left|\begin{array}{lll} \log x & p & 1 \\ \log y & q & 1 \\ \log z & r & 1 \end{array}\right|$ equals

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