If `a ,b ,c` are in G.P. with common ratio `r_1a n dalpha,beta,gamma` are in G.P. with common ratio `r_2` and equations `a x+alphay+z=0,b x+betay+z=0,c x+gammay+z=0` have only zero solution, then which of the following is not true?

Options

A. `r_(1) ne 1`
B. `r_(2) ne 1`
C. `r_(1) ne r_(2)`
D. none of these

Explanation

To solve the problem, we need to analyze the given conditions and equations step by step.

### Step 1: Understand the Given Information
We know that:
- \( a, b, c \) are in geometric progression (G.P.) with common ratio \( r_1 \).
- \( \alpha, \beta, \gamma \) are also in G.P. with common ratio \( r_2 \).
- The equations:
\[ ax + \alpha y + z = 0, \]
\[ bx + \beta y + z = 0, \]
\[ cx + \gamma y + z = 0 \]
have only the trivial solution (i.e., the only solution is \( x = 0, y = 0, z = 0 \)).

### Step 2: Express \( b \) and \( c \) in terms of \( a \)
Since \( a, b, c \) are in G.P. with common ratio \( r_1 \):
- \( b = ar_1 \)
- \( c = ar_1^2 \)

### Step 3: Express \( \beta \) and \( \gamma \) in terms of \( \alpha \)
Similarly, since \( \alpha, \beta, \gamma \) are in G.P. with common ratio \( r_2 \):
- \( \beta = \alpha r_2 \)
- \( \gamma = \alpha r_2^2 \)

### Step 4: Formulate the Determinant
The system of equations can be represented in matrix form as:
\[ \begin{vmatrix} a & \alpha & 1 \\ ar_1 & \alpha r_2 & 1 \\ ar_1^2 & \alpha r_2^2 & 1 \end{vmatrix} \]
For the system to have only the trivial solution, the determinant must be non-zero.

### Step 5: Calculate the Determinant
Calculating the determinant:
\[ D = \begin{vmatrix} a & \alpha & 1 \\ ar_1 & \alpha r_2 & 1 \\ ar_1^2 & \alpha r_2^2 & 1 \end{vmatrix} \]
We can factor out \( a \) from the first column and \( \alpha \) from the second column:
\[ D = a \alpha \begin{vmatrix} 1 & 1 & 1 \\ r_1 & r_2 & 1 \\ r_1^2 & r_2^2 & 1 \end{vmatrix} \]

### Step 6: Evaluate the Remaining Determinant
The remaining determinant can be evaluated using the formula for the determinant of a 3x3 matrix:
\[ \begin{vmatrix} 1 & 1 & 1 \\ r_1 & r_2 & 1 \\ r_1^2 & r_2^2 & 1 \end{vmatrix} = (r_1 - r_2)(r_1 - 1)(r_2 - 1) \]
Thus,
\[ D = a \alpha (r_1 - r_2)(r_1 - 1)(r_2 - 1) \]

### Step 7: Set Conditions for Non-Zero Determinant
For the determinant \( D \) to be non-zero:
- \( a \neq 0 \)
- \( \alpha \neq 0 \)
- \( r_1 - r_2 \neq 0 \)
- \( r_1 - 1 \neq 0 \)
- \( r_2 - 1 \neq 0 \)

### Step 8: Analyze the Options
The question asks which of the following statements is not true based on the conditions derived. Since we have established that \( r_1 \) and \( r_2 \) must be equal to 1 for the determinant to be zero, we can check the options against this condition.

### Conclusion
After evaluating the conditions, we find that if all conditions are satisfied, then all statements regarding \( r_1 \) and \( r_2 \) being equal to 1 and not equal to each other must be checked against the options provided.