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 analyze the given equations and the ordered triplets. The equations we have are:

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

The ordered triplets are given as \((1, 2, p)\), \((2, 8, -6)\), and \((\alpha^2 - 2\alpha, p, 1)\).

### Step 1: Set Up the Determinant

To find the values of \(a\), \(b\), and \(c\), we can use Cramer's rule. The determinant of the coefficients must be equal to zero for there to be multiple solutions. The determinant can be set up as follows:

\[ D = \begin{vmatrix} \frac{1}{a} & \frac{1}{b} & \frac{1}{c} \\ \frac{1}{b} & \frac{1}{c} & \frac{1}{a} \\ \frac{1}{c} & \frac{1}{a} & \frac{1}{b} \end{vmatrix} \]

### Step 2: Calculate the Determinant

Calculating the determinant \(D\):

\[ D = \frac{1}{abc} \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} - \text{(other terms)} \]

After performing the calculations, we find that:

\[ D = 3 \cdot \frac{1}{abc} - \left(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}\right) = 0 \]

This simplifies to:

\[ 3abc = \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \]

### Step 3: Substitute Values

Now we substitute the values from the triplets into the equations.

From the triplet \((2, 8, -6)\):

\[ 2 + 8 - 6 = a \implies a = 4 \]

From the triplet \((1, 2, p)\):

\[ 1 + 2 + p = 4 \implies p = 1 \]

From the triplet \((\alpha^2 - 2\alpha, p, 1)\):

\[ \alpha^2 - 2\alpha + p + 1 = 4 \implies \alpha^2 - 2\alpha + 1 + 1 - 4 = 0 \]

This simplifies to:

\[ \alpha^2 - 2\alpha - 2 = 0 \]

### Step 4: Solve the Quadratic Equation

We can solve the quadratic equation using the quadratic formula:

\[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]

This gives:

\[ \alpha = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3} \]

### Step 5: Find the Sum of All Values of Alpha

The sum of the roots \(1 + \sqrt{3}\) and \(1 - \sqrt{3}\) is:

\[ (1 + \sqrt{3}) + (1 - \sqrt{3}) = 2 \]

### Final Answer

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

---