If `(1,2, p), (2, 8, -6) and (alpha^(2)-2alpha,p,1)` are ordered triplet pair of the form `(x, y, z)` which satisfy all the equations `(x)/(a)+(y)/(b)+(z)/(c )=1, (x)/(b)+(y)/(c )+(z)/(a)=1 and (x)/(c )+(y)/(a)+(z)/(b)=1,` then the sum of all the values of `alpha` is equal to (where, `ab+bc+ca ne0`)

Options

A. 3
B. 2
C. 0
D. `-2`

Answer

To solve the problem, we need to find the values of \( \alpha \) such that the ordered triplets \( (1, 2, p) \), \( (2, 8, -6) \), and \( (\alpha^2 - 2\alpha, p, 1) \) satisfy the equations:

1. \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \)
2. \( \frac{x}{b} + \frac{y}{c} + \frac{z}{a} = 1 \)
3. \( \frac{x}{c} + \frac{y}{a} + \frac{z}{b} = 1 \)

where \( a, b, c \) are the sums of the respective coordinates of the triplets.

### Step 1: Determine the values of \( a, b, c \)

From the second triplet \( (2, 8, -6) \):
- \( a = 2 + 8 - 6 = 4 \)

### Step 2: Solve for \( p \)

Next, we substitute the first triplet \( (1, 2, p) \) into the equation:
- \( 1 + 2 + p = 4 \)
- Thus, \( p = 4 - 3 = 1 \)

### Step 3: Substitute \( p \) back into the third triplet

Now we have \( p = 1 \), and we substitute this into the third triplet:
- The third triplet becomes \( (\alpha^2 - 2\alpha, 1, 1) \).

### Step 4: Set up the equation for the third triplet

Now we can write the equation:
- \( \alpha^2 - 2\alpha + 1 + 1 + 1 = 4 \)
- Simplifying gives us:
\[ \alpha^2 - 2\alpha + 3 = 4 \]
- Rearranging gives:
\[ \alpha^2 - 2\alpha - 1 = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = -2, c = -1 \).
- The discriminant is:
\[ b^2 - 4ac = (-2)^2 - 4(1)(-1) = 4 + 4 = 8 \]
- Thus, the roots are:
\[ \alpha = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2} \]

### Step 6: Find the sum of the values of \( \alpha \)

The values of \( \alpha \) are \( 1 + \sqrt{2} \) and \( 1 - \sqrt{2} \). Therefore, the sum of the values of \( \alpha \) is:
\[ (1 + \sqrt{2}) + (1 - \sqrt{2}) = 2 \]

### Final Answer

The sum of all the values of \( \alpha \) is \( \boxed{2} \).