The value of `|alpha|` for which the system of equation
`alphax+y+z=alpha-1`
`x+alphay+z=alpha-1`
`x+y+alphaz=alpha-1`
has no solution , is `"____"`

Explanation

To find the value of \( |\alpha| \) for which the system of equations has no solution, we need to analyze the given equations:

1. \( \alpha x + y + z = \alpha - 1 \)
2. \( x + \alpha y + z = \alpha - 1 \)
3. \( x + y + \alpha z = \alpha - 1 \)

### Step 1: Write the system in matrix form
The system can be represented in matrix form as:

\[ \begin{bmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \alpha - 1 \\ \alpha - 1 \\ \alpha - 1 \end{bmatrix} \]

### Step 2: Determine the determinant of the coefficient matrix
Let \( D \) be the determinant of the coefficient matrix:

\[ D = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the determinant formula for a 3x3 matrix, we have:

\[ D = \alpha \begin{vmatrix} \alpha & 1 \\ 1 & \alpha \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \alpha \end{vmatrix} + 1 \begin{vmatrix} 1 & \alpha \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} \alpha & 1 \\ 1 & \alpha \end{vmatrix} = \alpha^2 - 1 \)

2. \( \begin{vmatrix} 1 & 1 \\ 1 & \alpha \end{vmatrix} = \alpha - 1 \)

3. \( \begin{vmatrix} 1 & \alpha \\ 1 & 1 \end{vmatrix} = 1 - \alpha \)

Substituting these back into the determinant:

\[ D = \alpha(\alpha^2 - 1) - (\alpha - 1) + (1 - \alpha) \]

Simplifying:

\[ D = \alpha^3 - \alpha - \alpha + 1 + 1 - \alpha \]
\[ D = \alpha^3 - 3\alpha + 2 \]

### Step 4: Set the determinant equal to zero
For the system to have no solution, we set \( D = 0 \):

\[ \alpha^3 - 3\alpha + 2 = 0 \]

### Step 5: Factor the polynomial
To find the roots, we can try possible rational roots. Testing \( \alpha = 1 \):

\[ 1^3 - 3(1) + 2 = 0 \]

Thus, \( \alpha = 1 \) is a root. We can factor the polynomial as:

\[ (\alpha - 1)(\alpha^2 + \alpha - 2) = 0 \]

Next, we factor \( \alpha^2 + \alpha - 2 \):

\[ \alpha^2 + \alpha - 2 = (\alpha - 1)(\alpha + 2) \]

Thus, the complete factorization is:

\[ (\alpha - 1)^2(\alpha + 2) = 0 \]

### Step 6: Find the roots
Setting each factor to zero gives:

1. \( \alpha - 1 = 0 \) → \( \alpha = 1 \)
2. \( \alpha + 2 = 0 \) → \( \alpha = -2 \)

### Step 7: Determine valid solutions
Since the system has no solution when \( \alpha = 1 \), we discard this value as it is not valid. Thus, the only value of \( \alpha \) for which the system has no solution is:

\[ \alpha = -2 \]

### Final Answer
The value of \( |\alpha| \) is:

\[ |\alpha| = 2 \]