A) \[(0,\,-2,\,2)\]
B) \[(-2,\,2,\,2)\]
C) \[(2,\,0,\,-2)\]
D) \[(2,\,2,\,0)\]
Correct Answer: C
Solution :
Since the required plane is containing the line \[\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z-1}{3}\]and its projection on the plane \[2x+3y-z=5.\] \[\therefore \]The normal vector of the required plane is \[\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & -1 & 3 \\ 2 & 3 & -1 \\ \end{matrix} \right|=-8\hat{i}+8\hat{j}+8\hat{k}\] So, direction ratio of normal is (\[-1,\,\,1,\,\,1\]) \[\therefore \]The equation of required plane is \[-(x-3)+(y+2)+(z-1)=0\] \[\Rightarrow \]\[-x+y+z+4=0\] Only point (\[2,\,\,0,\,\,-2\]) is satisfying the above equation.
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