If `alpha,beta` roots of `x^(2)-6p_(1)x+2=0,beta,gamma` are roots of `x^(2)-6p_(2)x+3=0and gamma,alpha` are roots of equation `x^(2)-6p_(3)x+6=0` where `p_(1),p_(2),p_(3)` are positive then
The values of `alpha, beta, gamma` respectively are

Options

A. `2,3,1`
B. `2,1,3`
C. `1,2,3`
D. `-1,-2,-3`

Answer

To solve the problem, we need to analyze the three quadratic equations given and find the values of the roots \( \alpha, \beta, \) and \( \gamma \).

### Step 1: Write down the equations and their relationships
We have three equations:
1. \( x^2 - 6p_1 x + 2 = 0 \) with roots \( \alpha \) and \( \beta \)
2. \( x^2 - 6p_2 x + 3 = 0 \) with roots \( \beta \) and \( \gamma \)
3. \( x^2 - 6p_3 x + 6 = 0 \) with roots \( \gamma \) and \( \alpha \)

From Vieta's formulas, we can derive the following relationships:
- For the first equation:
- Sum of roots: \( \alpha + \beta = 6p_1 \) (Equation 1)
- Product of roots: \( \alpha \beta = 2 \) (Equation 2)

- For the second equation:
- Sum of roots: \( \beta + \gamma = 6p_2 \) (Equation 3)
- Product of roots: \( \beta \gamma = 3 \) (Equation 4)

- For the third equation:
- Sum of roots: \( \gamma + \alpha = 6p_3 \) (Equation 5)
- Product of roots: \( \gamma \alpha = 6 \) (Equation 6)

### Step 2: Express \( \beta \) and \( \gamma \) in terms of \( \alpha \)
From Equation 2, we have:
\[ \beta = \frac{2}{\alpha} \]

Substituting \( \beta \) into Equation 4:
\[ \frac{2}{\alpha} \gamma = 3 \implies \gamma = \frac{3\alpha}{2} \]

### Step 3: Substitute \( \gamma \) back into the equations
Now substitute \( \gamma \) into Equation 6:
\[ \left(\frac{3\alpha}{2}\right) \alpha = 6 \implies \frac{3\alpha^2}{2} = 6 \implies 3\alpha^2 = 12 \implies \alpha^2 = 4 \implies \alpha = 2 \text{ or } \alpha = -2 \]

### Step 4: Determine \( \beta \) and \( \gamma \)
1. If \( \alpha = 2 \):
- \( \beta = \frac{2}{2} = 1 \)
- \( \gamma = \frac{3 \cdot 2}{2} = 3 \)

2. If \( \alpha = -2 \):
- \( \beta = \frac{2}{-2} = -1 \)
- \( \gamma = \frac{3 \cdot (-2)}{2} = -3 \)

### Step 5: Check the positivity condition
Since \( p_1, p_2, p_3 \) are all positive, we need \( \alpha + \beta + \gamma \) to be positive:
- For \( \alpha = 2, \beta = 1, \gamma = 3 \):
\[ \alpha + \beta + \gamma = 2 + 1 + 3 = 6 \text{ (positive)} \]
- For \( \alpha = -2, \beta = -1, \gamma = -3 \):
\[ \alpha + \beta + \gamma = -2 - 1 - 3 = -6 \text{ (not positive)} \]

### Conclusion
The only valid solution is:
\[ \alpha = 2, \beta = 1, \gamma = 3 \]

### Final Answer
The values of \( \alpha, \beta, \gamma \) are \( 2, 1, 3 \) respectively.

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