A) \[a+b+c=0\]
B) \[b+c=a\]
C) \[c+a=b\]
D) \[a+b=c\]
Correct Answer: A
Solution :
Given three straight lines \[ax+by-c=0\], \[bx+cy-a=0\], \[cx+ay-b=0\] are collinear, Then \[\left| \,\begin{matrix} a & b & -c \\ b & c & -a \\ c & a & -b \\ \end{matrix}\, \right|\,=\,0\]\[\Rightarrow \]\[-(a+b+c)\,\left| \,\begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \\ \end{matrix}\, \right|=0\] Clearly,\[(a+b+c)=0\]
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