If p + q + r = a + b + c = 0, then the determinant `|{:(pa,qb,rc),(qc,ra,pb),(rb,pc,qa):}|` equals

Options

A. 0
B. 1
C. pa + qb + rc
D. pa + qb + rc + a + b + c

Explanation

To solve the determinant \( D = \begin{vmatrix} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{vmatrix} \) given that \( p + q + r = 0 \) and \( a + b + c = 0 \), we can follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{vmatrix} \]

### Step 2: Expand the Determinant
We can expand the determinant using the first row:
\[ D = pa \begin{vmatrix} ra & pb \\ pc & qa \end{vmatrix} - qb \begin{vmatrix} qc & pb \\ rb & qa \end{vmatrix} + rc \begin{vmatrix} qc & ra \\ rb & pc \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
Now we calculate the 2x2 determinants one by one.

1. For \( \begin{vmatrix} ra & pb \\ pc & qa \end{vmatrix} \):
\[ = (ra)(qa) - (pb)(pc) = raqa - pbpc \]

2. For \( \begin{vmatrix} qc & pb \\ rb & qa \end{vmatrix} \):
\[ = (qc)(qa) - (pb)(rb) = qcqa - pbrb \]

3. For \( \begin{vmatrix} qc & ra \\ rb & pc \end{vmatrix} \):
\[ = (qc)(pc) - (ra)(rb) = qcpc - rarb \]

### Step 4: Substitute Back into the Determinant
Now substitute these back into the expression for \( D \):
\[ D = pa(raqa - pbpc) - qb(qcqa - pbrb) + rc(qcpc - rarb) \]

### Step 5: Simplify the Expression
Now we simplify the expression:
\[ D = paraqa - papbpc - qbqcqa + qbpbrb + rcqcpc - rcarb \]

### Step 6: Use the Given Conditions
Using the conditions \( p + q + r = 0 \) and \( a + b + c = 0 \), we can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Since \( a + b + c = 0 \), we have:
\[ a^3 + b^3 + c^3 = 3abc \]
Similarly, we can apply this to \( p, q, r \):
\[ p^3 + q^3 + r^3 = 3pqr \]

### Step 7: Final Result
Substituting these identities into our expression for \( D \):
\[ D = pqr(3abc) - abc(3pqr) = 3pqrabc - 3pqrabc = 0 \]

Thus, the value of the determinant is:
\[ \boxed{0} \]