evaluate: `|(a-b-c,2a,2a),(2b,b-c-a,2b),(2c,2c,c-a-b)|`

Explanation

To evaluate the determinant

\[ D = \begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} \]

we will follow a series of transformations to simplify the determinant.

### Step 1: Apply Row Transformation
We will first transform the first row \( R_1 \) by adding the second row \( R_2 \) and the third row \( R_3 \) to it. This gives us:

\[ R_1 \rightarrow R_1 + R_2 + R_3 \]

Calculating this, we have:

- First element: \( (a-b-c) + 2b + 2c = a + b + c \)
- Second element: \( 2a + (b-c-a) + 2c = a + b + 3c \)
- Third element: \( 2a + 2b + (c-a-b) = a + b + c \)

So the new determinant becomes:

\[ D = \begin{vmatrix} a+b+c & a+b+3c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Now, we can factor out \( a+b+c \) from the first row:

\[ D = (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} \]

### Step 3: Apply Column Transformation
Next, we will perform column transformations to simplify the determinant further. We will transform \( C_1 \) by subtracting \( C_2 \) and \( C_3 \):

\[ C_1 \rightarrow C_1 - C_2 \]
\[ C_2 \rightarrow C_2 - C_3 \]

This gives us:

\[ D = (a+b+c) \begin{vmatrix} 1-1 & 1 & 1-1 \\ 2b - (b-c-a) & b-c-a - 2b & 2b \\ 2c - 2c & 2c - (c-a-b) & c-a-b \end{vmatrix} \]

Calculating the new elements:

- First row becomes \( (0, 1, 0) \)
- Second row becomes \( (2b - (b-c-a), -b+c+a, 2b) = (b+c-a, -b+c+a, 2b) \)
- Third row becomes \( (0, 2c - (c-a-b), c-a-b) = (0, a+b+c, c-a-b) \)

So we have:

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

### Step 4: Expand the Determinant
Now we can expand the determinant using the first row:

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

Calculating this 2x2 determinant:

\[ = (b+c-a)(c-a-b) \]

Thus, we have:

\[ D = (a+b+c) \cdot (b+c-a)(c-a-b) \]

### Final Result
The final result is:

\[ D = (a+b+c)^3 \]