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 `p_(1),p_(2),p_(3)` respectively are

Options

A. `1/2,2/3,5/6`
B. `1,2,5`
C. `6,1,4`
D. `2,3/2,6/5`

Answer

To solve the problem, we need to find the values of \( p_1, p_2, \) and \( p_3 \) based on the given quadratic equations and their roots. Let's break it down step by step.

### Step 1: Analyze the first equation
The first equation is:
\[ x^2 - 6p_1 x + 2 = 0 \]
Let the roots be \( \alpha \) and \( \beta \).

From Vieta's formulas:
- The sum of the roots \( \alpha + \beta = 6p_1 \) (Equation 1)
- The product of the roots \( \alpha \beta = 2 \) (Equation 2)

### Step 2: Analyze the second equation
The second equation is:
\[ x^2 - 6p_2 x + 3 = 0 \]
Let the roots be \( \beta \) and \( \gamma \).

From Vieta's formulas:
- The sum of the roots \( \beta + \gamma = 6p_2 \) (Equation 3)
- The product of the roots \( \beta \gamma = 3 \) (Equation 4)

### Step 3: Analyze the third equation
The third equation is:
\[ x^2 - 6p_3 x + 6 = 0 \]
Let the roots be \( \gamma \) and \( \alpha \).

From Vieta's formulas:
- The sum of the roots \( \gamma + \alpha = 6p_3 \) (Equation 5)
- The product of the roots \( \gamma \alpha = 6 \) (Equation 6)

### Step 4: Express \( \beta \) in terms of \( \alpha \)
From Equation 2, we have:
\[ \beta = \frac{2}{\alpha} \]

### Step 5: Substitute \( \beta \) into Equation 4
Substituting \( \beta \) into Equation 4:
\[ \frac{2}{\alpha} \cdot \gamma = 3 \]
From this, we can express \( \gamma \):
\[ \gamma = \frac{3\alpha}{2} \]

### Step 6: Substitute \( \gamma \) into Equation 6
Now substituting \( \gamma \) into Equation 6:
\[ \left(\frac{3\alpha}{2}\right) \alpha = 6 \]
This simplifies to:
\[ \frac{3\alpha^2}{2} = 6 \]
Multiplying both sides by 2:
\[ 3\alpha^2 = 12 \]
Dividing by 3:
\[ \alpha^2 = 4 \]
Taking the square root:
\[ \alpha = 2 \quad (\text{since } p_1, p_2, p_3 \text{ are positive}) \]

### Step 7: Find \( \beta \) and \( \gamma \)
Using \( \alpha = 2 \):
\[ \beta = \frac{2}{\alpha} = \frac{2}{2} = 1 \]
\[ \gamma = \frac{3\alpha}{2} = \frac{3 \cdot 2}{2} = 3 \]

### Step 8: Calculate \( p_1, p_2, p_3 \)
Now we can find \( p_1, p_2, \) and \( p_3 \):
1. From Equation 1:
\[ \alpha + \beta = 2 + 1 = 3 = 6p_1 \implies p_1 = \frac{3}{6} = \frac{1}{2} \]
2. From Equation 3:
\[ \beta + \gamma = 1 + 3 = 4 = 6p_2 \implies p_2 = \frac{4}{6} = \frac{2}{3} \]
3. From Equation 5:
\[ \gamma + \alpha = 3 + 2 = 5 = 6p_3 \implies p_3 = \frac{5}{6} \]

### Final Values
Thus, the values of \( p_1, p_2, p_3 \) are:
\[ p_1 = \frac{1}{2}, \quad p_2 = \frac{2}{3}, \quad p_3 = \frac{5}{6} \]