1. Since $a_1,a_2,a_3,\dots$ are in G.P., let $a_k=Ar^{k-1}$ for some constants $A$ and $r$.

2. Taking logarithm, $\log a_k=\log A+(k-1)\log r.$ So $\log a_k$ is an arithmetic progression in $k$.

Let $\alpha=\log A, \qquad \beta=\log r.$ Then $\log a_k=\alpha+(k-1)\beta.$

3. Now write the entries of the determinant: [ \Delta= \begin{vmatrix} \log a_n & \log a_{n+1} & \log a_{n+2}\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5}\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{vmatrix}. ]

Using $\log a_k=\alpha+(k-1)\beta$, [ \Delta= \begin{vmatrix} \alpha+(n-1)\beta & \alpha+n\beta & \alpha+(n+1)\beta\ \alpha+(n+2)\beta & \alpha+(n+3)\beta & \alpha+(n+4)\beta\ \alpha+(n+5)\beta & \alpha+(n+6)\beta & \alpha+(n+7)\beta \end{vmatrix}. ]

4. Observe the rows form an arithmetic progression row-wise:

• Row 2 $-$ Row 1 $=(3\beta,3\beta,3\beta)$
• Row 3 $-$ Row 2 $=(3\beta,3\beta,3\beta)$

Hence, $R_3-R_2=R_2-R_1,$ so $R_1-2R_2+R_3=0.$ Thus the three rows are linearly dependent.

Therefore, the determinant must be $\Delta=0.$

5. Checking options:

• A: $1$ ❌
• B: $0$ ✅
• C: $4$ ❌
• D: $2$ ❌

Hence the correct answer is B.