If `|{:(,a+b+2c,a,b),(,c,b+c+2a,b),(,c,a,c+a+2b):}|=k(alphaa+betab+gammac)^(3)"then"(2alpha+beta-gamma)^(k)"is"(alpha,beta,gamma,k in Z^(+))`

Answer

To solve the given problem, we need to find the value of \( (2\alpha + \beta - \gamma)^k \) given the determinant condition. Let's break down the solution step-by-step.

### Step 1: Write down the determinant
We have the determinant:

\[ \begin{vmatrix} a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b \end{vmatrix} \]

### Step 2: Simplify the determinant
We can simplify the determinant by performing row operations. Specifically, we can subtract the first row from the second and third rows:

- \( R_2 \rightarrow R_2 - R_1 \)
- \( R_3 \rightarrow R_3 - R_1 \)

This gives us:

\[ \begin{vmatrix} a + b + 2c & a & b \\ c - (a + b + 2c) & (b + c + 2a) - a & b - b \\ c - (a + b + 2c) & a - (a + b + 2c) & (c + a + 2b) - b \end{vmatrix} \]

### Step 3: Calculate the new rows
After performing the row operations, we have:

- For \( R_2 \):
- First element: \( c - (a + b + 2c) = -a - b - c \)
- Second element: \( (b + c + 2a) - a = b + c + a \)
- Third element: \( 0 \)

- For \( R_3 \):
- First element: \( c - (a + b + 2c) = -a - b - c \)
- Second element: \( a - (a + b + 2c) = -b - 2c \)
- Third element: \( (c + a + 2b) - b = c + a + b \)

Thus, the determinant simplifies to:

\[ \begin{vmatrix} a + b + 2c & a & b \\ -a - b - c & b + c + a & 0 \\ -a - b - c & -b - 2c & c + a + b \end{vmatrix} \]

### Step 4: Factor out common terms
Notice that the first column has a common factor. We can factor out \( -1 \) from the second and third rows:

\[ = -1 \cdot \begin{vmatrix} a + b + 2c & a & b \\ a + b + c & b + c + a & 0 \\ a + b + c & -b - 2c & c + a + b \end{vmatrix} \]

### Step 5: Evaluate the determinant
Now we can evaluate the determinant. The determinant can be computed as:

\[ = (a + b + 2c) \cdot \begin{vmatrix} b + c + a & 0 \\ -b - 2c & c + a + b \end{vmatrix} \]

This determinant evaluates to:

\[ = (a + b + 2c) \cdot ((b + c + a)(c + a + b)) \]

### Step 6: Set the determinant equal to \( k(\alpha a + \beta b + \gamma c)^3 \)
Given that:

\[ |D| = k(\alpha a + \beta b + \gamma c)^3 \]

From the structure of the determinant, we can deduce that \( k = 2 \), \( \alpha = 1 \), \( \beta = 1 \), and \( \gamma = 1 \).

### Step 7: Calculate \( (2\alpha + \beta - \gamma)^k \)
Now we need to compute \( (2\alpha + \beta - \gamma)^k \):

\[ 2\alpha + \beta - \gamma = 2(1) + 1 - 1 = 2 \]

Thus, we have:

\[ (2\alpha + \beta - \gamma)^k = 2^2 = 4 \]

### Final Answer
The final answer is:

\[ \boxed{4} \]