If `alpha`, `beta`, `gamma` are roots of the equation `x^(2)(px+q)=r(x+1)`, then the value of determinant `|{:(1+alpha,1,1),(1, 1+beta,1),(1,1,1+gamma):}|` is

Options

A. `alphabetagamma`
B. `1+(1)/(alpha)+(1)/(beta)+(1)/(gamma)`
C. `0`
D. none of these

Explanation

To solve the problem, we need to find the value of the determinant given that \( \alpha, \beta, \gamma \) are the roots of the equation \( x^2(px + q) = r(x + 1) \).

### Step-by-Step Solution:

1. **Rearranging the Equation**:
Start with the equation \( x^2(px + q) = r(x + 1) \). Rearranging gives us:
\[ x^2(px + q) - r(x + 1) = 0 \]
This simplifies to:
\[ px^3 + qx^2 - rx - r = 0 \]

2. **Identifying the Roots**:
The roots of this polynomial are \( \alpha, \beta, \gamma \). By Vieta's formulas, we can express the sums and products of the roots:
- Sum of the roots: \( \alpha + \beta + \gamma = -\frac{q}{p} \)
- Sum of the products of the roots taken two at a time: \( \alpha\beta + \beta\gamma + \gamma\alpha = -\frac{r}{p} \)
- Product of the roots: \( \alpha\beta\gamma = \frac{r}{p} \)

3. **Setting Up the Determinant**:
The determinant we need to evaluate is:
\[ D = \begin{vmatrix} 1 + \alpha & 1 & 1 \\ 1 & 1 + \beta & 1 \\ 1 & 1 & 1 + \gamma \end{vmatrix} \]

4. **Column Operations**:
We can simplify the determinant by performing column operations. Divide the first column by \( \alpha \), the second column by \( \beta \), and the third column by \( \gamma \):
\[ D = \alpha\beta\gamma \begin{vmatrix} \frac{1 + \alpha}{\alpha} & \frac{1}{\alpha} & \frac{1}{\alpha} \\ \frac{1}{\beta} & \frac{1 + \beta}{\beta} & \frac{1}{\beta} \\ \frac{1}{\gamma} & \frac{1}{\gamma} & \frac{1 + \gamma}{\gamma} \end{vmatrix} \]

5. **Simplifying the Determinant**:
The determinant simplifies to:
\[ D = \alpha\beta\gamma \begin{vmatrix} 1 + \frac{1}{\alpha} & \frac{1}{\alpha} & \frac{1}{\alpha} \\ \frac{1}{\beta} & 1 + \frac{1}{\beta} & \frac{1}{\beta} \\ \frac{1}{\gamma} & \frac{1}{\gamma} & 1 + \frac{1}{\gamma} \end{vmatrix} \]

6. **Row Operations**:
Now, perform row operations to simplify further:
- Replace \( R_1 \) with \( R_1 - R_3 \) and \( R_2 \) with \( R_2 - R_3 \):
\[ D = \alpha\beta\gamma \begin{vmatrix} 0 & 0 & 0 \\ 0 & 1 + \frac{1}{\beta} - 1 & \frac{1}{\beta} - \frac{1}{\gamma} \\ \frac{1}{\gamma} & \frac{1}{\gamma} & 1 + \frac{1}{\gamma} \end{vmatrix} \]

7. **Evaluating the Determinant**:
The first row becomes all zeros, which means the determinant evaluates to zero:
\[ D = 0 \]

### Final Answer:
Thus, the value of the determinant is:
\[ \boxed{0} \]