The value of the determinant `Delta=|(logx,logy,logz),(log2x,log2y,log2z),(log3x,log3y,log3z)|` is

Options

A. 0
B. `log(xyz)`
C. `log(6xyz)`
D. `6log (xyz)`

Answer

To find the value of the determinant

\[ \Delta = \begin{vmatrix} \log x & \log y & \log z \\ \log 2x & \log 2y & \log 2z \\ \log 3x & \log 3y & \log 3z \end{vmatrix} \]

we can simplify the determinant using properties of logarithms.

### Step 1: Apply the logarithm property
Using the property of logarithms, \(\log(ab) = \log a + \log b\), we can rewrite the second and third rows:

- For the second row:
\[ \log 2x = \log 2 + \log x, \quad \log 2y = \log 2 + \log y, \quad \log 2z = \log 2 + \log z \]
So, the second row becomes:
\[ \begin{pmatrix} \log 2 + \log x & \log 2 + \log y & \log 2 + \log z \end{pmatrix} \]

- For the third row:
\[ \log 3x = \log 3 + \log x, \quad \log 3y = \log 3 + \log y, \quad \log 3z = \log 3 + \log z \]
So, the third row becomes:
\[ \begin{pmatrix} \log 3 + \log x & \log 3 + \log y & \log 3 + \log z \end{pmatrix} \]

Thus, we can rewrite the determinant as:

\[ \Delta = \begin{vmatrix} \log x & \log y & \log z \\ \log 2 + \log x & \log 2 + \log y & \log 2 + \log z \\ \log 3 + \log x & \log 3 + \log y & \log 3 + \log z \end{vmatrix} \]

### Step 2: Perform row operations
Next, we can simplify the determinant by performing row operations. We will subtract the first row from the second and third rows:

- New second row:
\[ (\log 2 + \log x - \log x, \log 2 + \log y - \log y, \log 2 + \log z - \log z) = (\log 2, \log 2, \log 2) \]

- New third row:
\[ (\log 3 + \log x - \log x, \log 3 + \log y - \log y, \log 3 + \log z - \log z) = (\log 3, \log 3, \log 3) \]

So, the determinant now looks like:

\[ \Delta = \begin{vmatrix} \log x & \log y & \log z \\ \log 2 & \log 2 & \log 2 \\ \log 3 & \log 3 & \log 3 \end{vmatrix} \]

### Step 3: Factor out common terms
Now, we can factor out the common terms from the second and third rows:

\[ \Delta = \log 2 \cdot \log 3 \cdot \begin{vmatrix} \log x & \log y & \log z \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 4: Evaluate the determinant
Notice that the second and third rows are identical. Therefore, the determinant of a matrix with two identical rows is zero:

\[ \begin{vmatrix} \log x & \log y & \log z \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0 \]

### Final Result
Thus, we have:

\[ \Delta = \log 2 \cdot \log 3 \cdot 0 = 0 \]

Therefore, the value of the determinant \(\Delta\) is

\[ \boxed{0} \]