A) 0
B) \[pa+qb+rc\]
C) 1
D) None of these
Correct Answer: A
Solution :
We have \[\left| \,\begin{matrix} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \\ \end{matrix}\, \right|\] \[=pqr({{a}^{3}}+{{b}^{3}}+{{c}^{3}})-abc({{p}^{3}}+{{q}^{3}}+{{r}^{3}})\] = \[pqr(3abc)-abc(3pqr)=0\], \[\left( \begin{align} & \because \,p+q+r=0\,,\,\therefore \,\,{{p}^{3}}+{{q}^{3}}+{{r}^{3}}=3pqr \\ & \because \,\,a+b+c=0\,,\therefore \,{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc \\ \end{align} \right)\].
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