Let `alpha,beta "and " gamma` are three distinct roots of
`|{:(x-1,-6,2),(-6,x-2,-4),(2,-4,x-6):}|` =0 the value of `((1)/(alpha)+(1)/(beta)+(1)/(gamma))^(-1)` is

Answer

To solve the problem, we need to find the value of \( \left( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \right)^{-1} \) where \( \alpha, \beta, \gamma \) are the distinct roots of the determinant equation given by:

\[ \begin{vmatrix} x - 1 & -6 & 2 \\ -6 & x - 2 & -4 \\ 2 & -4 & x - 6 \end{vmatrix} = 0 \]

### Step 1: Calculate the Determinant

We will calculate the determinant using the formula for a 3x3 matrix:

\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where \( a, b, c, d, e, f, g, h, i \) are the elements of the matrix.

Here, we can expand the determinant as follows:

\[ \text{Det} = (x - 1) \begin{vmatrix} x - 2 & -4 \\ -4 & x - 6 \end{vmatrix} - (-6) \begin{vmatrix} -6 & -4 \\ 2 & x - 6 \end{vmatrix} + 2 \begin{vmatrix} -6 & x - 2 \\ 2 & -4 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} x - 2 & -4 \\ -4 & x - 6 \end{vmatrix} = (x - 2)(x - 6) - (-4)(-4) = (x^2 - 8x + 12 - 16) = x^2 - 8x - 4 \)

2. \( \begin{vmatrix} -6 & -4 \\ 2 & x - 6 \end{vmatrix} = (-6)(x - 6) - (-4)(2) = -6x + 36 + 8 = -6x + 44 \)

3. \( \begin{vmatrix} -6 & x - 2 \\ 2 & -4 \end{vmatrix} = (-6)(-4) - (x - 2)(2) = 24 - (2x - 4) = 28 - 2x \)

Now substituting back into the determinant:

\[ \text{Det} = (x - 1)(x^2 - 8x - 4) + 6(-6x + 44) + 2(28 - 2x) \]

### Step 2: Expand and Simplify

Expanding the determinant:

\[ = (x^3 - 8x^2 - 4x - x^2 + 8x + 4) + (-36x + 264) + (56 - 4x) \]

Combining like terms:

\[ = x^3 - 9x^2 - 36x + 324 = 0 \]

### Step 3: Find the Roots

Using Vieta's formulas, we know:

- \( \alpha + \beta + \gamma = -\frac{-9}{1} = 9 \)
- \( \alpha\beta + \beta\gamma + \gamma\alpha = -\frac{-36}{1} = 36 \)
- \( \alpha\beta\gamma = -\frac{-324}{1} = 324 \)

### Step 4: Calculate \( \left( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \right)^{-1} \)

We know that:

\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} \]

Substituting the values we found:

\[ = \frac{36}{324} = \frac{1}{9} \]

Thus,

\[ \left( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \right)^{-1} = 9 \]

### Final Answer

The value of \( \left( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \right)^{-1} \) is \( \boxed{9} \).