A) \[-\frac{1}{2}\]
B) \[-\frac{1}{3}\]
C) \[-\frac{1}{6}\]
D) \[\frac{1}{6}\]
Correct Answer: C
Solution :
Given \[[a\,\,b\,\,c]\ne 0,\] i.e., non-coplanar. \[=a.\left\{ \frac{a\times c}{3b.(c\times a)} \right\}-b.\left\{ \frac{c\times a}{2c.(a\times c)} \right\}\] \[=\frac{a.(b\times c)}{3b.(c\times a)}-\frac{b.(c\times a)}{2c.(a\times b)}\] \[=\frac{[a\,b\,c]}{3\,[b\,c\,a]}-\frac{[b\,c\,a]}{2[c\,a\,b]}\,\,(\because a.(b\times b)=[a\,b\,c],\] \[[b\,c\,a]=[a\,b\,c],[c\,a\,b]=[b\,c\,a]\] \[=\frac{[a\,b\,c]}{3[a\,b\,c]}-\frac{[b\,c\,a]}{2[b\,c\,a]}\] \[=1/3-1/2=-1/6\]
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