If `lim_(xrarroo) xlog_(e)(|(alpha//x,1,gamma),(0,1//x,beta),(1,0,1//x)|)=-5.` where `alpha, beta, gamma` are finite real numbers, then

Options

A. `alpha=2, beta=1, gamma in R`
B. `alpha=2, beta=2, gamma=5`
C. `alpha in R, beta =1, gamma in R`
D. `alpha in R, beta = 1, gamma=5`

Explanation

To solve the limit problem given in the question, we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant given in the problem:
\[ D = \begin{vmatrix} \frac{\alpha}{x} & 1 & \gamma \\ 0 & \frac{1}{x} & \beta \\ 1 & 0 & \frac{1}{x} \end{vmatrix} \]

### Step 2: Calculate the Determinant
Using the formula for the determinant of a 3x3 matrix, we have:
\[ D = \frac{\alpha}{x} \left( \frac{1}{x} \cdot \frac{1}{x} - 0 \cdot 0 \right) - 1 \left( 0 \cdot \frac{1}{x} - \beta \cdot 1 \right) + \gamma \left( 0 \cdot 0 - 1 \cdot \frac{1}{x} \right) \]
Calculating this gives:
\[ D = \frac{\alpha}{x^3} + \beta - \frac{\gamma}{x} \]

### Step 3: Substitute the Determinant into the Limit
Now we substitute \(D\) into the limit expression:
\[ \lim_{x \to \infty} x \log\left(D\right) = \lim_{x \to \infty} x \log\left(\beta + \frac{\alpha}{x^3} - \frac{\gamma}{x}\right) \]

### Step 4: Analyze the Limit as \(x \to \infty\)
As \(x\) approaches infinity, \(\frac{\alpha}{x^3} \to 0\) and \(\frac{\gamma}{x} \to 0\). Therefore, we have:
\[ \lim_{x \to \infty} x \log\left(\beta\right) \]
We need this limit to equal \(-5\).

### Step 5: Determine Conditions for the Limit
For the limit to exist and be equal to \(-5\), we require:
1. \(\beta = 1\) (since \(\log(1) = 0\) gives us the indeterminate form \( \infty \cdot 0\)).
2. Substitute \(\beta = 1\) back into the limit:
\[ \lim_{x \to \infty} x \log\left(1 + \frac{\alpha}{x^3} - \frac{\gamma}{x}\right) \]

### Step 6: Use the Expansion for Logarithm
Using the logarithmic expansion \(\log(1 + u) \approx u\) for small \(u\):
\[ \lim_{x \to \infty} x \left(\frac{\alpha}{x^3} - \frac{\gamma}{x}\right) = \lim_{x \to \infty} \left(\frac{\alpha}{x^2} - \gamma\right) \]
As \(x \to \infty\), \(\frac{\alpha}{x^2} \to 0\), so we have:
\[ -\gamma = -5 \implies \gamma = 5 \]

### Step 7: Conclusion for \(\alpha\)
Since \(\alpha\) does not affect the limit, it can take any real value. Thus, we conclude:
- \(\alpha \in \mathbb{R}\)
- \(\beta = 1\)
- \(\gamma = 5\)

### Final Answer
The values are:
- \(\alpha \in \mathbb{R}\)
- \(\beta = 1\)
- \(\gamma = 5\)