Let $$a_1, a_2, a_3, \ldots, a_{10}$$ be in G.P. with $$a_i > 0$$ for $$i = 1, 2, \ldots, 10$$ and $$S$$ be the set of pairs $$(r, k)$$, $$r, k \in N$$ (the set of natural numbers) for which $$\begin{vmatrix} \log_e a_1^r a_2^k & \log_e a_2^r a_3^k & \log_e a_3^r a_4^k \\ \log_e a_4^r a_5^k & \log_e a_5^r a_6^k & \log_e a_6^r a_7^k \\ \log_e a_7^r a_8^k & \log_e a_8^r a_9^k & \log_e a_9^r a_{10}^k \end{vmatrix} = 0$$. Then the number of elements in $$S$$, is:

Let the given geometric progression have first term $$a_1=A$$ and common ratio $$q$$, where, as stated, every $$a_i>0$$. Hence, for every natural number $$n$$, we have the standard G.P. relation

$$a_n=Aq^{\,n-1}.$$

We need logarithms of these terms, so we write

$$\log_e a_n=\log_e\!\left(Aq^{\,n-1}\right)=\log_e A+\left(n-1\right)\log_e q.$$

For simplicity, put

$$\alpha=\log_e A, \qquad d=\log_e q.$$

Because the logarithms form an arithmetic progression, we may list them as

$$L_1=\alpha,\; L_2=\alpha+d,\;L_3=\alpha+2d,\;\ldots,\;L_{10}=\alpha+9d,$$

where, in general,

$$L_i=\alpha+\left(i-1\right)d.$$

Now consider the three-by-three matrix inside the determinant. Every entry is of the form $$\log_e a_j^{\,r}a_{j+1}^{\,k}$$ with suitable $$j$$. Using the logarithm law $$\log_e xy=\log_e x+\log_e y$$ together with $$\log_e x^r=r\log_e x$$, we get

$$\log_e a_j^{\,r}a_{j+1}^{\,k}=r\log_e a_j+k\log_e a_{j+1}=rL_j+kL_{j+1}.$$

Writing out all nine entries produces

$$M=\begin{pmatrix} rL_1+kL_2 & rL_2+kL_3 & rL_3+kL_4\\[4pt] rL_4+kL_5 & rL_5+kL_6 & rL_6+kL_7\\[4pt] rL_7+kL_8 & rL_8+kL_9 & rL_9+kL_{10} \end{pmatrix}.$$

Observe that each entry is linear in $$r$$ and $$k$$, so we can separate the whole matrix into a linear combination of two fixed matrices:

$$M=rP+kQ,$$

where

$$P=\begin{pmatrix} L_1 & L_2 & L_3\\ L_4 & L_5 & L_6\\ L_7 & L_8 & L_9 \end{pmatrix}, \qquad Q=\begin{pmatrix} L_2 & L_3 & L_4\\ L_5 & L_6 & L_7\\ L_8 & L_9 & L_{10} \end{pmatrix}.$$

Our determinant is therefore

$$\det M=\det\!\left(rP+kQ\right).$$

To decide when this determinant vanishes, we first explore the structure of $$P$$. Substitute each $$L_i=\alpha+(i-1)d$$ explicitly:

$$P=\begin{pmatrix} \alpha & \alpha+d & \alpha+2d\\ \alpha+3d & \alpha+4d & \alpha+5d\\ \alpha+6d & \alpha+7d & \alpha+8d \end{pmatrix}.$$

Look at the three row-vectors of $$P$$:

Row 1: $$\bigl(\alpha,\;\alpha+d,\;\alpha+2d\bigr)$$

Row 2: $$\bigl(\alpha+3d,\;\alpha+4d,\;\alpha+5d\bigr)=\text{Row}_1+3d\,(1,1,1)$$

Row 3: $$\bigl(\alpha+6d,\;\alpha+7d,\;\alpha+8d\bigr)=\text{Row}_1+6d\,(1,1,1).$$

Thus each row differs from the previous one by the same vector $$3d\,(1,1,1)$$. Consequently, the three rows are linearly dependent; in fact

$$\text{Row}_3-2\text{Row}_2+\text{Row}_1=0.$$

Linear dependence of the rows implies

$$\det P=0,$$

so $$\operatorname{rank}P\le 2$$.

The matrix $$Q$$ is obtained from $$P$$ by adding $$d\,(1,1,1)$$ to each column, i.e.

Row 1 of $$Q$$ is $$\text{Row}_1+d\,(1,1,1),$$

Row 2 of $$Q$$ is $$\text{Row}_2+d\,(1,1,1),$$

Row 3 of $$Q$$ is $$\text{Row}_3+d\,(1,1,1).$$

Therefore every row of $$Q$$ lies in the two-dimensional subspace spanned by

$$v_1=\bigl(\alpha,\;\alpha+d,\;\alpha+2d\bigr), \qquad v_2=(1,1,1).$$

Hence $$\operatorname{rank}Q\le 2$$ and so

$$\det Q=0.$$

Now compare the row spaces of $$P$$ and $$Q$$. Because both are contained in the span of $$v_1$$ and $$v_2$$, they are the same two-dimensional subspace. Consequently, any linear combination of $$P$$ and $$Q$$, i.e. $$rP+kQ$$, has all three rows lying in that same plane. This guarantees that

$$\operatorname{rank}\!\left(rP+kQ\right)\le 2\quad\Longrightarrow\quad\det\!\left(rP+kQ\right)=0$$

for every choice of the natural numbers $$r$$ and $$k$$ (indeed, for every real $$r,k$$).

Thus the determinant in the problem statement vanishes for all ordered pairs $$(r,k)\in\mathbb N\times\mathbb N$$. The set

$$S=\bigl\{(r,k)\in\mathbb N\times\mathbb N:\det M=0\bigr\}$$

therefore contains every such pair, and because $$\mathbb N\times\mathbb N$$ is infinite, so is $$S$$.

Hence, the correct answer is Option A.