If a, b, c are non zero complex numbers satisfying `a^(2) + b^(2) + c^(2) = 0 and |(b^(2) + c^(2),ab,ac),(ab,c^(2) + a^(2),bc),(ac,bc,a^(2) + b^(2))| = k a^(2) b^(2) c^(2)`, then k is equal to

Options

A. 3
B. 2
C. 4
D. 1

Explanation

To solve the problem, we need to evaluate the determinant given the conditions on the complex numbers \( a, b, c \).

### Step 1: Understanding the Given Condition
We have the condition:
\[ a^2 + b^2 + c^2 = 0 \]
This implies that \( a^2, b^2, c^2 \) are the roots of the polynomial \( x^3 + px + q = 0 \) where \( p = 0 \) and \( q = 0 \). This means that \( a^2, b^2, c^2 \) can be expressed in terms of each other.

### Step 2: Setting Up the Determinant
We need to evaluate the determinant:
\[ D = \begin{vmatrix} b^2 + c^2 & ab & ac \\ ab & c^2 + a^2 & bc \\ ac & bc & a^2 + b^2 \end{vmatrix} \]

### Step 3: Simplifying the Determinant
To simplify the determinant, we can perform row operations. We can multiply the first row by \( a \), the second row by \( b \), and the third row by \( c \):
\[ D = abc \begin{vmatrix} b^2 + c^2 & ab & ac \\ ab & c^2 + a^2 & bc \\ ac & bc & a^2 + b^2 \end{vmatrix} \]

### Step 4: Factor Out Common Terms
Notice that each column has a common factor:
- Column 1 has \( b^2 + c^2 \)
- Column 2 has \( ab \)
- Column 3 has \( ac \)

We can factor these out:
\[ D = abc \cdot \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 5: Evaluating the Determinant
Now, we can evaluate the determinant:
\[ D = 2 \cdot \begin{vmatrix} b^2 + c^2 & 0 & 0 \\ 0 & c^2 + a^2 & 0 \\ 0 & 0 & a^2 + b^2 \end{vmatrix} \]
This determinant simplifies to:
\[ D = 2(b^2 + c^2)(c^2 + a^2)(a^2 + b^2) \]

### Step 6: Relating to \( k a^2 b^2 c^2 \)
We know from the problem statement that:
\[ D = k a^2 b^2 c^2 \]
Setting the two expressions for \( D \) equal gives:
\[ 2(b^2 + c^2)(c^2 + a^2)(a^2 + b^2) = k a^2 b^2 c^2 \]

### Step 7: Finding \( k \)
To find \( k \), we need to evaluate the left-hand side under the condition \( a^2 + b^2 + c^2 = 0 \). Using this condition, we can express \( b^2 + c^2 = -a^2 \), \( c^2 + a^2 = -b^2 \), and \( a^2 + b^2 = -c^2 \).

Substituting these into the equation gives:
\[ 2(-a^2)(-b^2)(-c^2) = k a^2 b^2 c^2 \]
This simplifies to:
\[ 2 a^2 b^2 c^2 = k a^2 b^2 c^2 \]
Dividing both sides by \( a^2 b^2 c^2 \) (since \( a, b, c \) are non-zero):
\[ k = 2 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{4} \]