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

Options

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)`

Answer

d

Explanation

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)`