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Question

Let a=^i+^j+^k,b=2^i+^k,c=^i+2^j+3^k, then the ascending order the following is
A.[a×b b×c c×a]
B.[a+b b+c c+a]
C.[abc][a1 b1 c1]
D.[ab bc ca]

A
B,D,C,A
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B
D,A,B,C
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C
A,B,C,D
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D
C,A,B,D
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Solution

The correct option is A B,D,C,A
We now,
a×b=∣ ∣ ∣^i^j^ka1a2a3b1b2b3∣ ∣ ∣
For
A.[a×b b×c c×a]
here a×b=^i+^j2^k,
b×c=2^i5^j+4^k,
c×a=^i+2^j^k
[a×b b×c c×a]=∣ ∣112254121∣ ∣=9
for
B.[a+b b+c c+a]=∣ ∣312324234∣ ∣=6
for
C.[abc][a1 b1 c1]=1 and
for
D.[ab bc ca]=∣ ∣110122012∣ ∣=0
So, ascending order is : B,D,C,A

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