A) \[-1\]
B) 1
C) 2
D) 0
Correct Answer: C
Solution :
Key Idea: If three points are collinear, then area of triangle should be zero. Since, the points \[(p\,+1,1),(2p\,+1,3)\] and \[(2p+2,2p)\] are collinear, then \[\left| \begin{matrix} p+1 & 1 & 1 \\ 2p+1 & 3 & 1 \\ 2p+2 & 2p & 1 \\ \end{matrix} \right|=0\] Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\]and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[\Rightarrow \]\[\left| \begin{matrix} p+1 & 1 & 1 \\ p & 2 & 0 \\ p+1 & 2p-1 & 0 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[1(2{{p}^{2}}-p-2p-2)=0\] \[\Rightarrow \]\[2{{p}^{2}}-3p-2=0\] \[\Rightarrow \]\[(2p+1)(p-2)=0\Rightarrow p=2,-\frac{1}{2}\]
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