If `|{:(a,b,aalpha-b),(b,c,balpha-c),(2,1,0):}|=0`, then

Options

A. a,b,c are in A.P.
B. a,b,c are in G.P.
C. `alpha=2`
D. none of these

Answer

To solve the problem given by the determinant equation:

\[ \begin{vmatrix} a & b & a\alpha - b \\ b & c & b\alpha - c \\ 2 & 1 & 0 \end{vmatrix} = 0 \]

we will follow these steps:

### Step 1: Replace the third column
We will replace the third column with a linear combination of the first two columns. Specifically, we will replace the third column with:

\[ \text{Column 3} - \alpha \cdot \text{Column 1} + \text{Column 2} \]

This gives us the new third column as:

\[ \begin{pmatrix} a\alpha - b - \alpha a + b \\ b\alpha - c - \alpha b + c \\ 0 - 2\alpha + 1 \end{pmatrix} \]

### Step 2: Simplify the new determinant
After performing the column operation, the determinant becomes:

\[ \begin{vmatrix} a & b & 0 \\ b & c & 0 \\ 2 & 1 & 1 - 2\alpha \end{vmatrix} \]

### Step 3: Expand the determinant
Now we can expand this determinant along the third column:

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

The determinant of the 2x2 matrix is:

\[ \begin{vmatrix} a & b \\ b & c \end{vmatrix} = ac - b^2 \]

Thus, we have:

\[ (1 - 2\alpha)(ac - b^2) = 0 \]

### Step 4: Set up the equations
From the equation above, we have two cases:

1. \(1 - 2\alpha = 0\) which gives \(\alpha = \frac{1}{2}\)
2. \(ac - b^2 = 0\) which gives \(ac = b^2\)

### Step 5: Conclusion
From the second case, we can conclude that \(a\), \(b\), and \(c\) are in geometric progression (G.P.) since:

\[ b^2 = ac \implies \frac{b}{a} = \frac{c}{b} \]

Thus, the relationship between \(a\), \(b\), and \(c\) is that they are in G.P.

### Final Answer
The relation between \(a\), \(b\), and \(c\) is that they are in geometric progression (G.P.).

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