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If the system of equations, ax+by+cz=0 , bx+cy+az=0  and cx+ay+bz=0  have non trivial solution, then

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a
a+b+c=0
b
a=b=c
c
Either a+b+c=0  ort a=b=c
d
None of these

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detailed solution

Correct option is A

For a non-trivial solution, abcbcacab=0 ⇒(a+b+c)1bc1ca1ab=0⇒(a+b+c)1bc0c−ba−c0a−bb−c=0⇒(a+b+c)×(−a2−b2−c2+ab+bc+ca)=0⇒−12(a+b+c){(a−b)2+(b−c)2+(c−a)2}=0⇒  Either a+b+c=0Or (a−b)2+(b−c)2+(c−a)2=0⇒  Either a+b+c=0Or a−b=0 , b−c=0  and c−a=0 .

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