Answer:
It is given that \[\vec{\alpha },\] \[\vec{\beta }\] and \[\vec{\gamma }\] are coplanar vectors. \[\therefore \] \[[\vec{\alpha }\,\,\vec{\beta }\,\,\vec{\gamma }]=0\] \[\Rightarrow \] \[\left| \begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[abc-a-c+1+1-b=0\] \[\Rightarrow \] \[abc=a+b+c-2\] ?. (i) Now, \[\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\] \[=\frac{(1-b)(1-c)+(1-c)(1-a)+(1-a)(1-b)}{(1-a)(1-b)(1-c)}\] \[=\frac{3-2(a\,+b+c)+(ab\,+bc\,+ca)}{1-(a\,+b\,+c)+(ab\,+bc\,+ca)-abc}\] \[=\frac{3-2(a\,+b\,+c)+ab\,+bc\,+ca}{1-(a\,+b\,+c)+(ab\,+bc\,+ca)-(a\,+b\,+c-2)}\] [Using Eq. (i)] \[=\frac{3-2(a\,+b\,+c)+ab\,+bc\,+ca}{3-2(a\,+b\,+c)+ab\,+bc\,+ca}=1\] Hence proved.
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