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Q.

If x2+y2+z2≠0 , x=cy+bz,y=az+cx and z=bx+ay  then a2+b2+c2+2abc=

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a

0

b

1

c

2

d

–1

answer is B.

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Detailed Solution

|1−c−bc−1aba−1 | = 0 ⇒1−a2−c2−abc−abc−b2=0 ⇒ a2+b2+c2+2abc=1
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