Question:

Step 1: Express the terms of the geometric progression. Let the first term of the geometric progression (GP) be $A$ and the common ratio be $R$. The given terms $x, y, z, p, q, r, l, m, n$ are consecutive terms in this GP. We can write them as: $x = A$ $y = AR$ $z = AR^2$ $p = AR^3$ $q = AR^4$ $r = AR^5$ $l = AR^6$ $m = AR^7$ $n = AR^8$ Step 2: Convert the terms into their logarithmic form. Taking the logarithm of each term (to any base), we use the properties $\log(xy) = \log x + \log y$ and $\log(x^k) = k \log x$. Let $\log A = a_0$ and $\log R = d_0$. Then the logarithmic terms form an arithmetic progression (AP): $\log x = \log A = a_0$ $\log y = \log(AR) = \log A + \log R = a_0 + d_0$ $\log z = \log(AR^2) = \log A + 2\log R = a_0 + 2d_0$ $\log p = \log(AR^3) = \log A + 3\log R = a_0 + 3d_0$ $\log q = \log(AR^4) = \log A + 4\log R = a_0 + 4d_0$ $\log r = \log(AR^5) = \log A + 5\log R = a_0 + 5d_0$ $\log l = \log(AR^6) = \log A + 6\log R = a_0 + 6d_0$ $\log m = \log(AR^7) = \log A + 7\log R = a_0 + 7d_0$ $\log n = \log(AR^8) = \log A + 8\log R = a_0 + 8d_0$ Step 3: Substitute the logarithmic expressions into the determinant. The given determinant is: $$ \Delta = \begin{vmatrix} \log x & \log y & \log z \\ \log p & \log q & \log r \\ \log l & \log m & \log n \end{vmatrix} $$ Substituting the expressions from Step 2: $$ \Delta = \begin{vmatrix} a_0 & a_0 + d_0 & a_0 + 2d_0 \\ a_0 + 3d_0 & a_0 + 4d_0 & a_0 + 5d_0 \\ a_0 + 6d_0 & a_0 + 7d_0 & a_0 + 8d_0 \end{vmatrix} $$ Step 4: Identify the relationship between the rows of the determinant. Let $R_1$, $R_2$, and $R_3$ denote the first, second, and third rows of the determinant, respectively. $R_1 = [a_0, a_0 + d_0, a_0 + 2d_0]$ $R_2 = [a_0 + 3d_0, a_0 + 4d_0, a_0 + 5d_0]$ $R_3 = [a_0 + 6d_0, a_0 + 7d_0, a_0 + 8d_0]$ Observe the difference between consecutive rows: $R_2 - R_1 = [(a_0 + 3d_0) - a_0, (a_0 + 4d_0) - (a_0 + d_0), (a_0 + 5d_0) - (a_0 + 2d_0)]$ $R_2 - R_1 = [3d_0, 3d_0, 3d_0]$ $R_3 - R_2 = [(a_0 + 6d_0) - (a_0 + 3d_0), (a_0 + 7d_0) - (a_0 + 4d_0), (a_0 + 8d_0) - (a_0 + 5d_0)]$ $R_3 - R_2 = [3d_0, 3d_0, 3d_0]$ Since $R_2 - R_1 = R_3 - R_2$, it follows that $R_3 = 2R_2 - R_1$. This equation shows that the rows are linearly dependent. Step 5: Conclude the value of the determinant. A fundamental property of determinants states that if the rows (or columns) of a matrix are linearly dependent, the value of the determinant is zero. Since $R_3 - 2R_2 + R_1 = 0$, the rows of the determinant are linearly dependent. Therefore, the value of the determinant is $0$. The final answer is $\boxed{0}$.