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 the steps outlined in the video transcript.

### Step 1: Modify the first row
We will replace the first row \( R_1 \) with \( R_1 + R_2 + R_3 \).

\[ R_1 = (a-b-c) + (2b) + (2c), \quad R_2 = (2b, b-c-a, 2b), \quad R_3 = (2c, 2c, c-a-b) \]

Calculating \( R_1 \):

\[ R_1 = (a-b-c + 2b + 2c) = (a + b + c) \]

So, the determinant becomes:

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

### Step 2: Simplify the determinant
Next, we will perform column operations. We will replace \( C_2 \) with \( C_2 - C_1 \) and \( C_3 \) with \( C_3 - C_1 \).

Calculating the new columns:

- \( C_2 = 2a - (a + b + c) = a - b - c \)
- \( C_3 = 2a - (a + b + c) = a - b - c \)

Thus, the determinant becomes:

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

This simplifies to:

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

### Step 3: Factor out common terms
Notice that the second and third columns are identical. Therefore, the determinant evaluates to zero:

\[ D = 0 \]

### Final Answer
The value of the determinant is

\[ \boxed{0} \]