A) 0
B) 1
C) 2
D) - 1
Correct Answer: D
Solution :
Since, \[\vec{a}=\hat{i}+a\hat{j}+{{a}^{2}}\hat{k},\vec{b}=\hat{i}+b\hat{j}+{{b}^{2}}\hat{k}\] and \[\vec{c}=\hat{i}+c\hat{j}+{{c}^{2}}\hat{k}\] are coplanar vectors \[\therefore \] \[[\vec{a}\,\vec{b}\,\vec{c}]\ne 0\] \[\Rightarrow \] \[\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|\ne 0\] \[\Rightarrow \] \[\Delta \ne 0\] where \[\Delta =\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|\] Now, \[\left| \begin{matrix} a & {{a}^{2}} & 1+{{a}^{3}} \\ b & {{b}^{2}} & 1+{{b}^{3}} \\ c & {{c}^{2}} & 1+{{c}^{3}} \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|+\left| \begin{matrix} a & {{a}^{2}} & {{a}^{3}} \\ b & {{b}^{2}} & {{b}^{3}} \\ c & {{c}^{2}} & {{c}^{3}} \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|+abc\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\Delta +abc\,\Delta =0\] \[\Rightarrow \] \[\Delta (1+abc)=0\] \[\Rightarrow \] \[1+abc=0\] \[(\because \,\Delta \ne 0)\] \[\Rightarrow \] \[abc=-1\]
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