if `x ne 0 , y ne 0 ,z ne 0 " and " |{:(1+x,,1,,1),(1+y,,1+2y,,1),(1+z,,1+z,,1+3z):}|=0` then
`x^(-1) +y^(-1) +z^(-1)` is equal to

Options

A. `-1`
B. `-2`
C. `-3`
D. none of these

Explanation

To solve the given problem, we need to evaluate the determinant and find the value of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \).

Given the determinant:
\[ D = \begin{vmatrix} 1 + x & 1 & 1 \\ 1 + y & 1 + 2y & 1 \\ 1 + z & 1 + z & 1 + 3z \end{vmatrix} \]
We know that \( |D| = 0 \).

### Step 1: Factor out \( xyz \)
First, we can factor out \( xyz \) from the determinant. We rewrite the determinant as follows:
\[ D = xyz \begin{vmatrix} \frac{1}{x} + 1 & 1 & 1 \\ \frac{1}{y} + 1 & \frac{1 + 2y}{y} & 1 \\ \frac{1}{z} + 1 & \frac{1 + z}{z} & \frac{1 + 3z}{z} \end{vmatrix} \]
This simplifies to:
\[ D = xyz \begin{vmatrix} 1 + \frac{1}{x} & 1 & 1 \\ 1 + \frac{1}{y} & 1 + 2 & 1 \\ 1 + \frac{1}{z} & 1 + 1 & 1 + 3 \end{vmatrix} \]

### Step 2: Row operations
Now, we perform row operations to simplify the determinant. We can add the rows together:
\[ R_1 \rightarrow R_1 + R_2 + R_3 \]
This gives us:
\[ D = xyz \begin{vmatrix} 3 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} & 1 & 1 \\ 1 + \frac{1}{y} & 1 + 2 & 1 \\ 1 + \frac{1}{z} & 1 + 1 & 1 + 3 \end{vmatrix} \]

### Step 3: Expand the determinant
Now we can expand the determinant. The determinant is set to zero:
\[ xyz \left(3 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) = 0 \]

### Step 4: Analyze the equation
Since \( x, y, z \neq 0 \), we know that \( xyz \neq 0 \). Therefore, we can conclude:
\[ 3 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \]

### Step 5: Solve for \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \)
Rearranging the equation gives us:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -3 \]

Thus, the final answer is:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -3 \]