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Perpendiculars are drawn from points on the line `(x+2)/2=(y+1)/(-1)=z/3` to the plane `x + y + z=3` The feet of perpendiculars lie on the line

A

`(x)/(5)= (y-1)/(8) = (z-2)/(-13)`

B

`(x)/(2)= (y+1)/(3)=(z-2)/(-5)`

C

`(x)/(4)= (y-1)/(3)= (z-2)/(-7)`

D

`(x)/(2)= (y-1)/(-7)= (z-2)/(5)`

Text Solution

Verified by Experts

The correct Answer is:
d
Any point B on line is `(2lamda-2, -lamda-1, 3lamda)`
Point B lies on the plane for some `lamda`
`rArr" "(2lamda-2)+ (-lamda-1)+ 3lamda=3 or lamda = 3//2`
`rArr" "B-= (1, -5//2, 9//2)`
The foot of the perpendicular from the point `(-2, -1, 0)` on the plane is the point `A(0, 1, 2)`
`rArr" "` Direction ratio of `AB= (1, (-7)/(2), (5)/(2))-= (2, -7, 5)`.
Hence, feet of perpendicular lies on the line
`(x)/(2)= (y-1)/(-7)= (z-2)/(5)`
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