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. `pa+qb+rc`
C. 1
D. none of these

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 along the first row (R1):
\[ D = pa \begin{vmatrix} ra & pb \\ pc & qa \end{vmatrix} - qb \begin{vmatrix} qc & pb \\ rb & qa \end{vmatrix} + rc \begin{vmatrix} qc & ra \\ rb & pc \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
Calculating the first 2x2 determinant:
\[ \begin{vmatrix} ra & pb \\ pc & qa \end{vmatrix} = ra \cdot qa - pb \cdot pc = r a q a - p b c \]

Calculating the second 2x2 determinant:
\[ \begin{vmatrix} qc & pb \\ rb & qa \end{vmatrix} = qc \cdot qa - pb \cdot rb = q c q a - p b r b \]

Calculating the third 2x2 determinant:
\[ \begin{vmatrix} qc & ra \\ rb & pc \end{vmatrix} = qc \cdot pc - ra \cdot rb = q c p c - r a r b \]

### Step 4: Substitute Back into the Determinant
Substituting these back into the determinant:
\[ D = pa (r a q a - p b c) - qb (q c q a - p b r b) + rc (q c p c - r a r b) \]

### Step 5: Simplify the Expression
Now we simplify each term:
1. The first term becomes:
\[ pa (r a q a - p b c) = p a r a q a - p^2 a b c \]
2. The second term becomes:
\[ -qb (q c q a - p b r b) = -q^2 b c q a + p q b^2 r \]
3. The third term becomes:
\[ rc (q c p c - r a r b) = r c q c p c - r^2 c a r b \]

### Step 6: Combine Like Terms
Combining all these terms:
\[ D = p a r a q a - p^2 a b c - q^2 b c q a + p q b^2 r + r c q c p c - r^2 c a r b \]

### Step 7: Factor Out Common Terms
Notice that we can factor out \( pqr \) and \( abc \) from the expression:
\[ D = pqr (a^3 + b^3 + c^3) - abc (p^3 + q^3 + r^3) \]

### Step 8: Use the Given Conditions
Using the identities given in the problem:
Since \( a + b + c = 0 \) and \( p + q + r = 0 \), we can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = 0 \]
Thus, \( a^3 + b^3 + c^3 = 3abc \) and similarly \( p^3 + q^3 + r^3 = 3pqr \).

### Final Result
Substituting these back gives:
\[ D = pqr(3abc) - abc(3pqr) = 0 \]

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