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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

Any point `k` on the given line `(x+2)/2=(y+1)/-1=z/3` can be shown as `(2lambda-2,-lambda-1,3lambda)`
Equation of plane is `x+y+z=3`.
So, point k should satisfy plane equation.
Thus,
`2lambda - 2-lambda-1+3lambda = 3`
`=>4lambda = 6`
`=>lambda = 3/2`
Putting this value of `lambda`, point k becomes (1,-5/2,9/2). ...
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