The value of the determinant `|[log_a(x/y), log_a(y/z), log_a(z/x)], [log_b (y/z), log_b (z/x), log_b (x/y)], [log_c (z/x), log_c (x/y), log_c (y/z)]|`

Options

A. 1
B. -1
C. 0
D. `(1)/(6) log _(a) xyz`

Explanation

To solve the determinant

\[ D = \begin{vmatrix} \log_a\left(\frac{x}{y}\right) & \log_a\left(\frac{y}{z}\right) & \log_a\left(\frac{z}{x}\right) \\ \log_b\left(\frac{y}{z}\right) & \log_b\left(\frac{z}{x}\right) & \log_b\left(\frac{x}{y}\right) \\ \log_c\left(\frac{z}{x}\right) & \log_c\left(\frac{x}{y}\right) & \log_c\left(\frac{y}{z}\right) \end{vmatrix} \]

we will use properties of logarithms and determinants.

### Step 1: Rewrite the Determinant
We start by rewriting the determinant in a clearer form:

\[ D = \begin{vmatrix} \log_a(x) - \log_a(y) & \log_a(y) - \log_a(z) & \log_a(z) - \log_a(x) \\ \log_b(y) - \log_b(z) & \log_b(z) - \log_b(x) & \log_b(x) - \log_b(y) \\ \log_c(z) - \log_c(x) & \log_c(x) - \log_c(y) & \log_c(y) - \log_c(z) \end{vmatrix} \]

### Step 2: Apply the Property of Determinants
Using the property of determinants that allows us to replace a column by the sum of itself and other columns, we can add all three columns together to the first column:

\[ D = \begin{vmatrix} \log_a\left(\frac{x}{y}\right) + \log_a\left(\frac{y}{z}\right) + \log_a\left(\frac{z}{x}\right) & \log_a\left(\frac{y}{z}\right) & \log_a\left(\frac{z}{x}\right) \\ \log_b\left(\frac{y}{z}\right) + \log_b\left(\frac{z}{x}\right) + \log_b\left(\frac{x}{y}\right) & \log_b\left(\frac{z}{x}\right) & \log_b\left(\frac{x}{y}\right) \\ \log_c\left(\frac{z}{x}\right) + \log_c\left(\frac{x}{y}\right) + \log_c\left(\frac{y}{z}\right) & \log_c\left(\frac{x}{y}\right) & \log_c\left(\frac{y}{z}\right) \end{vmatrix} \]

### Step 3: Simplify Using Logarithmic Properties
Using the property of logarithms that states \(\log_m(a) + \log_m(b) + \log_m(c) = \log_m(abc)\), we can simplify the first column:

\[ D = \begin{vmatrix} \log_a\left(\frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x}\right) & \log_a\left(\frac{y}{z}\right) & \log_a\left(\frac{z}{x}\right) \\ \log_b\left(\frac{y}{z} \cdot \frac{z}{x} \cdot \frac{x}{y}\right) & \log_b\left(\frac{z}{x}\right) & \log_b\left(\frac{x}{y}\right) \\ \log_c\left(\frac{z}{x} \cdot \frac{x}{y} \cdot \frac{y}{z}\right) & \log_c\left(\frac{x}{y}\right) & \log_c\left(\frac{y}{z}\right) \end{vmatrix} \]

### Step 4: Evaluate the Logarithmic Products
Notice that:

\[ \frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x} = 1 \]

Thus, we have:

\[ D = \begin{vmatrix} \log_a(1) & \log_a\left(\frac{y}{z}\right) & \log_a\left(\frac{z}{x}\right) \\ \log_b(1) & \log_b\left(\frac{z}{x}\right) & \log_b\left(\frac{x}{y}\right) \\ \log_c(1) & \log_c\left(\frac{x}{y}\right) & \log_c\left(\frac{y}{z}\right) \end{vmatrix} \]

### Step 5: Substitute Values of Logarithms
Since \(\log_a(1) = 0\), \(\log_b(1) = 0\), and \(\log_c(1) = 0\), the first column becomes zero:

\[ D = \begin{vmatrix} 0 & \log_a\left(\frac{y}{z}\right) & \log_a\left(\frac{z}{x}\right) \\ 0 & \log_b\left(\frac{z}{x}\right) & \log_b\left(\frac{x}{y}\right) \\ 0 & \log_c\left(\frac{x}{y}\right) & \log_c\left(\frac{y}{z}\right) \end{vmatrix} \]

### Step 6: Conclude the Determinant Value
Since one entire column of the determinant is zero, we conclude that the value of the determinant is:

\[ D = 0 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{0} \]