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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

To find the foot of perpendiculars and find its locus.
Fromula used
Footof perpendicular from `(x_(1),y_(1),z_(1))`to
`ax+by+cz+d=0`be`(x_(2),y_(2),z_(2))`, then
`(x_(2)-x_(1))/(a)=(y_(2)-y_(1))/(b)=(z_(2)-z_(1))/(c)=(-(ax_(1)+by_(1)+cz_(1)+d))/(a^(2)+b^(2)+c^(2))`
Any point on `(x+2)/(2)=(y+1)/(1)=(z)/(3)=lambda`
`implies" "x=2lambda-2,y=-lambda-1,z=3lambda`
Let foot of perpendicular from `(2lambda-2,-lambda-1,3lambda)` to `x+y+z=3` be `(x_(2),y_(2),z_(2)).`
`:.(x_(2)-(2lambda-2))/(1)=(y_(2)-(-lambda-1))/(1)=(z_(2)-(3lambda))/(1)`
`=-((2lambda-2-lambda-1+3lambda-3))/(1+1+1)`
`implies" "x_(2)-2lambda+2=y_(2)+lambda+1=z_(2)-3lambda=2-(4lambda)/(3)`
`:." "x_(2)=(2lambda)/(3),y_(2)=1-(7lambda)/(3),z_(2)=2+(5lambda)/(3)`
`implies" "lambda=(x_(2)-0)/(2//3)=(y_(2)-1)/(-7//3)=(z_(2)-2)/(5//3)`
Hence, foot of perpendicular lie on
`(x)/(2//3)=(y-1)/(-7//3)=(z-2)/(5//3)implies(x)/(2)=(y-1)/(-7)=(z-2)/(5)`