If `|(a-b-c,2a,2a),(2b,b-c-a,2b),(2c,2c,c-a-b)|=k(a+b+c)^(3)`, then k is equal to

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the determinant problem given by the expression \( |(a-b-c, 2a, 2a), (2b, b-c-a, 2b), (2c, 2c, c-a-b)| = k(a+b+c)^3 \), we will follow these steps:

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

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

### Step 2: Simplify the First Row
We can simplify the first row by adding the second and third rows to it:

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

Calculating the new first row:

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

Thus, the determinant becomes:

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

### Step 3: Factor Out \( (a + b + c) \)
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 4: Simplify the Remaining Determinant
Now, we will perform column operations to simplify the remaining determinant:

- Let \( C_1 \rightarrow C_1 - C_2 \) and \( C_2 \rightarrow C_2 - C_3 \):

This gives us:

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

### Step 5: Evaluate the Determinant
The determinant now has a column of zeros, which means:

\[ D = 0 \]

This indicates that the determinant is zero unless \( a + b + c = 0 \).

### Step 6: Compare with Given Expression
Since we have \( D = k(a + b + c)^3 \), and we found \( D = 0 \), we can conclude that \( k \) must be equal to \( 0 \) when \( a + b + c \neq 0 \).

### Conclusion
Thus, the value of \( k \) is:

\[ \boxed{0} \]