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
If `A(alpha,(1)/(alpha)),B(beta,(1)/(beta)),C(gamma,(1)/(gamma))` be vertices of `DeltaABC` then centroid of `DeltaABC` is

Options

A. `(4,(11)/(18))`
B. `(2,(11)/(18))`
C. `(-2(11)/(18))`
D. `(-4(11)/(18))`

Answer

To solve the problem step by step, we will first analyze the given equations and find the values of the roots \( \alpha, \beta, \) and \( \gamma \). Then we will determine the vertices of triangle \( ABC \) and finally calculate the centroid.

### Step 1: Finding the roots \( \alpha, \beta, \gamma \)

1. **First Equation**:
\[ x^2 - 6p_1x + 2 = 0 \]
- Sum of roots \( \alpha + \beta = 6p_1 \)
- Product of roots \( \alpha \beta = 2 \)

2. **Second Equation**:
\[ x^2 - 6p_2x + 3 = 0 \]
- Sum of roots \( \beta + \gamma = 6p_2 \)
- Product of roots \( \beta \gamma = 3 \)

3. **Third Equation**:
\[ x^2 - 6p_3x + 6 = 0 \]
- Sum of roots \( \gamma + \alpha = 6p_3 \)
- Product of roots \( \gamma \alpha = 6 \)

### Step 2: Setting up equations

We have the following equations:
1. \( \alpha + \beta = 6p_1 \) and \( \alpha \beta = 2 \)
2. \( \beta + \gamma = 6p_2 \) and \( \beta \gamma = 3 \)
3. \( \gamma + \alpha = 6p_3 \) and \( \gamma \alpha = 6 \)

### Step 3: Expressing \( p_1, p_2, p_3 \)

From the product of the roots:
- From the first equation: \( \beta = \frac{2}{\alpha} \)
- From the second equation: \( \gamma = \frac{3}{\beta} = \frac{3\alpha}{2} \)
- From the third equation: \( \alpha = \frac{6}{\gamma} = \frac{6 \cdot 2}{3} = 4 \)

### Step 4: Solving for \( \beta \) and \( \gamma \)

1. Substitute \( \alpha = 4 \) into \( \alpha \beta = 2 \):
\[ 4\beta = 2 \implies \beta = \frac{1}{2} \]

2. Substitute \( \beta = \frac{1}{2} \) into \( \beta \gamma = 3 \):
\[ \frac{1}{2} \gamma = 3 \implies \gamma = 6 \]

### Step 5: Values of \( \alpha, \beta, \gamma \)

Now we have:
- \( \alpha = 4 \)
- \( \beta = \frac{1}{2} \)
- \( \gamma = 6 \)

### Step 6: Finding the vertices of triangle \( ABC \)

The vertices are given as:
- \( A(\alpha, \frac{1}{\alpha}) = (4, \frac{1}{4}) \)
- \( B(\beta, \frac{1}{\beta}) = \left(\frac{1}{2}, 2\right) \)
- \( C(\gamma, \frac{1}{\gamma}) = (6, \frac{1}{6}) \)

### Step 7: Calculating the centroid

The formula for the centroid \( G \) of triangle \( ABC \) is:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the coordinates:
\[ G\left(\frac{4 + \frac{1}{2} + 6}{3}, \frac{\frac{1}{4} + 2 + \frac{1}{6}}{3}\right) \]

Calculating the x-coordinate:
\[ \frac{4 + \frac{1}{2} + 6}{3} = \frac{10.5}{3} = 3.5 \]

Calculating the y-coordinate:
Finding a common denominator for \( \frac{1}{4}, 2, \frac{1}{6} \):
\[ \frac{1}{4} = \frac{3}{12}, \quad 2 = \frac{24}{12}, \quad \frac{1}{6} = \frac{2}{12} \]
\[ \frac{\frac{1}{4} + 2 + \frac{1}{6}}{3} = \frac{3 + 24 + 2}{12 \cdot 3} = \frac{29}{36} \]

Thus, the centroid \( G \) is:
\[ G\left(3.5, \frac{29}{36}\right) \]

### Final Answer
The centroid of triangle \( ABC \) is:
\[ \boxed{\left(3.5, \frac{29}{36}\right)} \]