A) -1
B) 0
C) 1
D) 4
Correct Answer: A
Solution :
since\[(1,a,{{a}^{2}})(1,b,{{b}^{2}})\]and\[(1,c,{{c}^{2}})\]are non-coplanar \[\therefore \] \[\Delta =\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|\ne 0\] and \[\left| \begin{matrix} a & {{a}^{2}} & 1+{{a}^{3}} \\ b & {{b}^{2}} & 1+{{b}^{3}} \\ c & {{c}^{2}} & 1+{{c}^{3}} \\ \end{matrix} \right|=\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|\] \[+\left| \begin{matrix} a & {{a}^{2}} & {{a}^{3}} \\ b & {{b}^{2}} & {{b}^{3}} \\ c & {{c}^{2}} & {{c}^{3}} \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\Delta +abc\Delta =0\] \[\Rightarrow \] \[\Delta (abc+1)=0\] \[\Rightarrow \] \[abc=-0\] \[(\because \Delta \ne 0)\]
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