A) \[{{t}_{1}}{{t}_{2}}{{t}_{3}}=1\]
B) \[{{t}_{1}}+{{t}_{2}}+{{t}_{3}}={{t}_{1}}{{t}_{2}}{{t}_{3}}\]
C) \[{{t}_{1}}+{{t}_{2}}+{{t}_{3}}=0\]
D) \[{{t}_{1}}+{{t}_{2}}+{{t}_{3}}=-1\]
Correct Answer: C
Solution :
The given points are collinear, if \[\left| \begin{matrix} {{t}_{1}} & 2a{{t}_{1}}+at_{1}^{3} & 1 \\ {{t}_{2}} & 2a{{t}_{2}}+at_{2}^{3} & 1 \\ {{t}_{3}} & 2a{{t}_{3}}+at_{3}^{3} & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[a\left| \begin{matrix} {{t}_{1}} & 2{{t}_{1}}+t_{1}^{3} & 1 \\ {{t}_{2}} & 2{{t}_{2}}+t_{2}^{3} & 1 \\ {{t}_{3}} & 2{{t}_{3}}+t_{3}^{3} & 1 \\ \end{matrix} \right|=0\] Applying\[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},\,\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\], we get \[\left| \begin{matrix} {{t}_{1}} & 2{{t}_{1}}+t_{1}^{3} & 1 \\ {{t}_{2}}-{{t}_{1}} & 2({{t}_{2}}-{{t}_{1}})+(t_{2}^{3}-t_{1}^{3}) & 0 \\ {{t}_{3}}-{{t}_{1}} & 2({{t}_{3}}-{{t}_{1}})+(t_{3}^{3}-t_{1}^{3}) & 0 \\ \end{matrix} \right|\] \[\Rightarrow \] \[({{t}_{2}}-{{t}_{1}})({{t}_{3}}-{{t}_{1}})\] \[\left| \begin{matrix} {{t}_{1}} & 2{{t}_{1}}+t_{1}^{3} & 1 \\ 1 & 2+t_{2}^{2}+t_{1}^{2}+{{t}_{2}}{{t}_{1}} & 0 \\ 1 & 2+t_{3}^{2}+t_{1}^{2}+{{t}_{3}}{{t}_{1}} & 0 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[({{t}_{2}}-{{t}_{1}})({{t}_{3}}-{{t}_{1}})({{t}_{3}}-{{t}_{2}})({{t}_{3}}+{{t}_{2}}+{{t}_{1}})=0\] \[\Rightarrow \]\[{{t}_{1}}+{{t}_{2}}+{{t}_{3}}=0\] \[[\because {{t}_{1}}\ne {{t}_{2}}\ne {{t}_{3}}]\]
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