If `alpha` is a non-real cube root of `-2`, then the value of `|(1,2 alpha,1),(alpha^(2),1,3 alpha^(2)),(2,2 alpha,1)|`, is

Options

A. `-11`
B. `-12`
C. `-13`
D. 0

Explanation

To solve the determinant \( D = \begin{vmatrix} 1 & 2\alpha & 1 \\ \alpha^2 & 1 & 3\alpha^2 \\ 2 & 2\alpha & 1 \end{vmatrix} \), where \( \alpha \) is a non-real cube root of \(-2\), we will follow these steps:

### Step 1: Write down the determinant
We start with the determinant:

\[ D = \begin{vmatrix} 1 & 2\alpha & 1 \\ \alpha^2 & 1 & 3\alpha^2 \\ 2 & 2\alpha & 1 \end{vmatrix} \]

### Step 2: Expand the determinant using the first row
We will expand the determinant using the first row. The formula for the determinant of a 3x3 matrix is:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix, we have:

- \( a = 1 \), \( b = 2\alpha \), \( c = 1 \)
- The remaining elements are:
- \( d = \alpha^2 \), \( e = 1 \), \( f = 3\alpha^2 \)
- \( g = 2 \), \( h = 2\alpha \), \( i = 1 \)

Thus, we calculate:

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

### Step 3: Calculate each part
Calculating each term:

1. First term:

\[ 1 \cdot (1 - 6\alpha^3) \]

2. Second term:

\[ - 2\alpha \cdot (\alpha^2 - 6\alpha^2) = -2\alpha \cdot (-5\alpha^2) = 10\alpha^3 \]

3. Third term:

\[ 2\alpha^3 - 2 \]

### Step 4: Combine all terms
Now we combine all the terms:

\[ D = (1 - 6\alpha^3) + 10\alpha^3 + (2\alpha^3 - 2) \]

Combining like terms:

\[ D = 1 - 2 - 6\alpha^3 + 10\alpha^3 + 2\alpha^3 \]

This simplifies to:

\[ D = -1 + 6\alpha^3 \]

### Step 5: Substitute the value of \( \alpha^3 \)
Since \( \alpha \) is a cube root of \(-2\), we have:

\[ \alpha^3 = -2 \]

Substituting this into our expression for \( D \):

\[ D = -1 + 6(-2) = -1 - 12 = -13 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{-13} \]