A) \[p+q+r=0\]
B) \[{{p}^{2}}+{{q}^{2}}+{{r}^{2}}=pq+qr+rp\]
C) \[{{p}^{3}}+{{q}^{3}}+{{r}^{3}}=3pqr\]
D) None of the above
Correct Answer: C
Solution :
Three lines\[px+qy+r=0,\,\,\,qx+ry+p=0\] and \[rx+py+q=0\] are concurrent, if\[\left| \begin{matrix} p & q & r \\ q & r & p \\ r & p & q \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[(p+q+r)({{p}^{2}}+{{q}^{2}}+{{r}^{2}}-pq-qr-rp)=0\] \[\Rightarrow \]\[p+q+r=0\]or\[{{p}^{2}}+{{q}^{2}}+{{r}^{2}}=pq+qr+rp\] ? (i) Now, \[{{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr=(p+q+r)\] \[({{p}^{2}}+{{q}^{2}}+{{r}^{2}}-pq-qr-rp)\] \[{{p}^{3}}+{{q}^{3}}+{{r}^{3}}=3pqr\] [From Eq. (i)]
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