A) \[\frac{x-2}{3}=\frac{y-1}{5}=\frac{z-1}{-1}\]
B) \[\frac{x-2}{2}=\frac{y-1}{-1}=\frac{z-1}{1}\]
C) \[\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z+1}{1}\]
D) \[\frac{x-1}{3}=\frac{y-3}{5}=\frac{z-5}{1}\]
Correct Answer: D
Solution :
[d] Let the direction ratios of the line be (a, b, c). Then \[\therefore \text{ }2a-b+c=0\]and \[a-b-2c=0\] Solving, we get \[\frac{a}{3}=\frac{b}{5}=\frac{c}{-1}\] Thus, the direction ratios of the line are \[(3,5,-1)\]. Any point on the line L is \[(2+\lambda ,2-\lambda ,3-2\lambda )\]. It lies on the plane P if \[2(2+\lambda )-(2-\lambda )+(3-2\lambda )=4\] \[\Rightarrow \,\,\,\,\,\,\,\,\lambda =-1\] So, the point of intersection of the line and the plane is \[(1,3,5)\]. Hence, the equation of the required line is: \[\frac{x-1}{3}=\frac{y-3}{5}=\frac{z-5}{1}\]
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