A) \[6\]
B) \[5\]
C) \[4\]
D) \[7\]
Correct Answer: D
Solution :
Let \[\overrightarrow{OA}=\hat{i}-\hat{j}+\hat{k},\] \[\overrightarrow{OB}=2\hat{i}+3\hat{j}+4\hat{k}\] and \[\overrightarrow{OC}=3\hat{i}+7\hat{j}+p\hat{k}\] Now, \[\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\hat{i}+4\hat{j}+3\hat{k}\] and \[\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=2\hat{i}+8\hat{j}+(p-1)\hat{k}\] If the points are collinear, then \[\overrightarrow{AB}=\lambda \overrightarrow{AC}\] \[\Rightarrow \] \[\hat{i}+4\hat{j}++3\hat{k}=\lambda [2\hat{i}+8\hat{j}+(p-1)\hat{k}]\] On comparing the coefficients of \[\hat{i},\,\hat{j},\hat{k}\] both sides, we get \[2\lambda =1\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\lambda =\frac{1}{2},\] and \[(p-1)\lambda =3\] \[\Rightarrow \] \[(p-1).\frac{1}{2}=3\] \[\Rightarrow \] \[p-1=6\,\,\,\,\Rightarrow \,\,p=7\]
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