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Question

If [abc]=2, then a.(b×c)(c×a).b+b.(c×a)(a×b).c+c.(a×b)(b×c).a=

A
3
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B
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C
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D
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Solution

The correct option is B 3
¯¯¯a.(¯¯bׯ¯c)(¯¯cׯ¯¯a).¯¯b+¯¯b.(¯¯cׯ¯¯a)(¯¯¯aׯ¯b).¯¯c+¯¯c.(¯¯¯aׯ¯b)(¯¯bׯ¯c).¯¯¯a
=(¯¯bׯ¯c)¯¯¯a(¯¯cׯ¯¯a).¯¯b+(¯¯cׯ¯¯a)¯¯b(¯¯¯aׯ¯b).¯¯c+(¯¯¯aׯ¯b)¯¯c(¯¯bׯ¯c).¯¯¯a, since uv=vu
=1+1+1=3, Since (¯¯¯aׯ¯b)¯¯c= (¯¯bׯ¯c)¯¯¯a=(¯¯cׯ¯¯a)¯¯b

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