the determinant `|{:(a,,b,,aalpha+b),(b,,c,,balpha+c),(aalpha+b,,balpha+c,,0):}|=0` is equal to zero if

Options

A. a,b,c are in A.P
B. a,b,c are in G.P.
C. `alpha` is a root of the equation `ax^(2) +bx+c=0`
D. `(x-alpha)` is a factor fo `ax^(2) +2bx+c`

Explanation

To solve the determinant equation

\[ \begin{vmatrix} a & b & a\alpha + b \\ b & c & b\alpha + c \\ a\alpha + b & b\alpha + c & 0 \end{vmatrix} = 0, \]

we will follow these steps:

### Step 1: Write the Determinant
We start by writing the determinant explicitly:

\[ D = \begin{vmatrix} a & b & a\alpha + b \\ b & c & b\alpha + c \\ a\alpha + b & b\alpha + c & 0 \end{vmatrix} \]

### Step 2: Apply Column Operations
We will perform column operations to simplify the determinant. Specifically, we will replace the third column \(C_3\) with \(C_3 - \alpha C_1 - C_2\):

\[ C_3 \rightarrow C_3 - \alpha C_1 - C_2 \]

This gives us:

\[ D = \begin{vmatrix} a & b & (a\alpha + b) - \alpha a - b \\ b & c & (b\alpha + c) - \alpha b - c \\ a\alpha + b & b\alpha + c & 0 - \alpha(a\alpha + b) - (b\alpha + c) \end{vmatrix} \]

### Step 3: Simplify the Third Column
Now we simplify each entry in the third column:

1. First row: \( (a\alpha + b) - \alpha a - b = 0 \)
2. Second row: \( (b\alpha + c) - \alpha b - c = 0 \)
3. Third row: \( 0 - \alpha(a\alpha + b) - (b\alpha + c) = -\alpha(a\alpha + b) - (b\alpha + c) \)

Thus, the determinant becomes:

\[ D = \begin{vmatrix} a & b & 0 \\ b & c & 0 \\ a\alpha + b & b\alpha + c & -\alpha(a\alpha + b) - (b\alpha + c) \end{vmatrix} \]

### Step 4: Factor Out Common Terms
Now, we can factor out the common terms from the determinant. We can take out \(-\alpha(a\alpha + 2b + c)\) from the third column:

\[ D = -\alpha(a\alpha + 2b + c) \begin{vmatrix} a & b & 1 \\ b & c & 1 \\ a\alpha + b & b\alpha + c & 1 \end{vmatrix} \]

### Step 5: Evaluate the Remaining Determinant
The remaining determinant can be evaluated using the properties of determinants. We can expand it, but we will focus on the conditions for it to equal zero.

### Step 6: Set the Determinant Equal to Zero
For the determinant \(D\) to be zero, either \(-\alpha(a\alpha + 2b + c) = 0\) or the remaining determinant must equal zero.

1. \(-\alpha = 0\) implies \(\alpha = 0\).
2. \(a\alpha + 2b + c = 0\) gives us a relationship between \(a\), \(b\), and \(c\).

### Step 7: Conclude with the Conditions
From the determinant being zero, we can conclude that:

- The values \(a\), \(b\), and \(c\) must satisfy \(b^2 = ac\) (indicating they are in geometric progression).
- The polynomial \(a\alpha^2 + 2b\alpha + c = 0\) implies that \(\alpha\) is a root of this quadratic equation.

### Final Result
Thus, the determinant is equal to zero if:

1. \(b^2 = ac\) (indicating \(a\), \(b\), \(c\) are in geometric progression).
2. \(\alpha\) is a root of the equation \(a\alpha^2 + 2b\alpha + c = 0\).