A) 0
B) 1
C) 2
D) 4
Correct Answer: A
Solution :
\[[a-b\,\,b-c\,\,c-a]=\{(a-b)\times (b-c)\}\,.\,(c-a)\] \[=(a\times b-a\times c-b\times b+b\times c)\,.\,(c-a)\] \[=(a\times ab+ca\times a+b\times c)\,.\,(\mathbf{c}-\mathbf{a})\] \[=(a\times b)\,.\,c-(a\times b)\,.\,a+(c\times a)\,.\,c-(c\times a)\,.\,a\] = \[(\mathbf{a}\times \mathbf{b}).\mathbf{c}-(\mathbf{a}\times \mathbf{b}).\mathbf{a}+(\mathbf{c}\times \mathbf{a}).\mathbf{c}-(\mathbf{c}\times \mathbf{a}).\mathbf{a}\] \[+(\mathbf{b}\times \mathbf{c}).\mathbf{c}-(\mathbf{b}\times \mathbf{c}).\mathbf{a}\] \[=[a\,b\,c]-[a\,b\,a]+[c\,a\,c]-[c\,a\,a]+[b\,c\,c]-[b\,c\,a]\]= 0.
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