If `|(a-b-c,2a,2a),(2b,b-c-a,2b),(2c,2c,c-a-b)|=(a+b+c)^(lamda)`, then `lamda =`

Options

A. 1
B. 2
C. 3
D. none

Answer

To solve the determinant problem given by

\[ \left| \begin{array}{ccc} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{array} \right| = (a+b+c)^{\lambda} \]

we will compute the determinant step by step.

### Step 1: Write down the determinant

We start with the determinant:

\[ D = \left| \begin{array}{ccc} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{array} \right| \]

### Step 2: Apply row operations

We can simplify the determinant by performing row operations. Let's add the first row to the second and third rows:

1. \( R_2 \to R_2 + R_1 \)
2. \( R_3 \to R_3 + R_1 \)

This gives us:

\[ D = \left| \begin{array}{ccc} a-b-c & 2a & 2a \\ 2b + (a-b-c) & (b-c-a) + 2a & 2b + 2a \\ 2c + (a-b-c) & 2c + 2a & (c-a-b) + 2a \end{array} \right| \]

Calculating the new rows:

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

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

So we have:

\[ D = \left| \begin{array}{ccc} a-b-c & 2a & 2a \\ a + b - c & b - c + a & 2(b + a) \\ a + c - b & 2(c + a) & c - b + a \end{array} \right| \]

### Step 3: Factor out common terms

We can factor out \( a + b + c \) from the first column. This gives us:

\[ D = (a + b + c) \left| \begin{array}{ccc} 1 & 2a & 2a \\ 1 & b - c + a & 2(b + a) \\ 1 & 2(c + a) & c - b + a \end{array} \right| \]

### Step 4: Simplify the determinant

Now we can perform row operations again. Subtract the first row from the second and third rows:

1. \( R_2 \to R_2 - R_1 \)
2. \( R_3 \to R_3 - R_1 \)

This gives us:

\[ D = (a + b + c) \left| \begin{array}{ccc} 1 & 2a & 2a \\ 0 & (b - c + a - 2a) & (2(b + a) - 2a) \\ 0 & (2(c + a) - 2a) & (c - b + a - 2a) \end{array} \right| \]

### Step 5: Calculate the determinant

Now we can simplify the determinant further. We can compute the determinant of the 2x2 matrix formed by the last two rows.

After simplifying, we find that the determinant can be expressed as:

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

### Step 6: Equate and find \( \lambda \)

Now we have:

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

Given that

\[ D = (a + b + c)^{\lambda} \]

we can equate the exponents:

\[ \lambda = 3 \]

### Final Answer

Thus, the value of \( \lambda \) is

\[ \boxed{3} \]