The system of equations
`ax+by+(a alpha+ b)z=0`
`bx+cy+(b alpha+c)z=0`
`(a alpha+b)x+(balpha+c)y=0`
has a non zero solutions if a,b,c are in

Options

A. A.P.
B. G.P.
C. H.P.
D. A.G.P

Answer

To solve the given system of equations for non-zero solutions, we need to analyze the determinant of the coefficient matrix. The equations are:

1. \( ax + by + (a\alpha + b)z = 0 \)
2. \( bx + cy + (b\alpha + c)z = 0 \)
3. \( (a\alpha + b)x + (b\alpha + c)y = 0 \)

### Step 1: Write the Coefficient Matrix
The coefficient matrix \( A \) for the system of equations can be represented as:

\[ A = \begin{bmatrix} a & b & a\alpha + b \\ b & c & b\alpha + c \\ a\alpha + b & b\alpha + c & 0 \end{bmatrix} \]

### Step 2: Calculate the Determinant
To find the conditions for non-zero solutions, we need to set the determinant of the coefficient matrix to zero:

\[ \text{det}(A) = 0 \]

### Step 3: Expand the Determinant
Using the determinant formula for a 3x3 matrix, we can expand it as follows:

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

### Step 4: Calculate Each Minor
Calculating the minors:

1. For the first minor:
\[ \begin{vmatrix} c & b\alpha + c \\ b\alpha + c & 0 \end{vmatrix} = c \cdot 0 - (b\alpha + c)(b\alpha + c) = - (b\alpha + c)^2 \]

2. For the second minor:
\[ \begin{vmatrix} b & b\alpha + c \\ a\alpha + b & 0 \end{vmatrix} = b \cdot 0 - (b\alpha + c)(a\alpha + b) = - (b\alpha + c)(a\alpha + b) \]

3. For the third minor:
\[ \begin{vmatrix} b & c \\ a\alpha + b & b\alpha + c \end{vmatrix} = b(b\alpha + c) - c(a\alpha + b) = b^2\alpha + bc - ac - bc = b^2\alpha - ac \]

### Step 5: Substitute Minors Back into the Determinant
Now substituting these back into the determinant equation:

\[ \text{det}(A) = a \cdot (-(b\alpha + c)^2) - b \cdot (-(b\alpha + c)(a\alpha + b)) + (a\alpha + b)(b^2\alpha - ac) \]

### Step 6: Set the Determinant to Zero
Setting the determinant to zero gives us:

\[ -a(b\alpha + c)^2 + b(b\alpha + c)(a\alpha + b) + (a\alpha + b)(b^2\alpha - ac) = 0 \]

### Step 7: Factor the Equation
This can be factored or simplified to find conditions on \( a, b, c \). The resulting conditions will lead us to determine if \( a, b, c \) are in Arithmetic Progression (AP), Geometric Progression (GP), Harmonic Progression (HP), or Arithmetic-Geometric Progression (AGP).

### Conclusion
After simplification, we find that the condition for non-zero solutions leads to:

- \( a\alpha^2 + 2b\alpha + c = 0 \) or \( ac - b^2 = 0 \)

The second condition \( ac - b^2 = 0 \) indicates that \( a, b, c \) are in GP.

Thus, the answer is that \( a, b, c \) are in **Geometric Progression (GP)**.

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