A) \[(3,\,\,-5,\,\,-3)\]
B) \[(4,\,\,-7,\,\,-9)\]
C) \[(0,\,\,2,\,\,-1)\]
D) \[(-3,\,\,5,\,\,3)\]
Correct Answer: B
Solution :
Given, equation of line is \[\frac{x-2}{1}=\frac{y+3}{-2}=\frac{z+5}{-2}\] DR's of given line is \[<1,\,\,-2,\,\,-2>\] Then, DC's of given line is\[<\frac{1}{3},\,\,\frac{-2}{3},\,\,\frac{-2}{3}>\] Thus, the given equation can be rewritten as \[\frac{x-2}{\frac{1}{3}}=\frac{y+3}{\frac{-2}{3}}=\frac{z+5}{-\frac{2}{3}}=r\] [say] \[\therefore \]Any point on the line is \[\left( 2+\frac{r}{3},\,\,3-\frac{2r}{3},\,\,-5-\frac{2r}{3} \right)\] where, \[r=\pm 6\] Hence, points are\[(4,\,\,-7,\,\,-9)\]and\[(0,\,\,1,\,\,-1)\].
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