Find the equation of the perpendicular drawn from `(2,4,-1)` to the line `(x+5)/1=(y+3)/4=(z-6)/3`.

Explanation

We have equation of the line as `(x+5)/1=(y+3)/4=(z-6)/(-9)=lamda`
`implies x=lamda-5,y=4lamda-3,z=6-9lamda`
Let the coordinate of `L` be `(lamda5,4lamda-3,6-9lamda),` then Dr's of PL are `(lamda-7,4lamda-7,7-9lamda)`
Also, the direction ratios of given line are proportional to `1,4,-9`.
Since PL is perpendicular to the given line.
`:. (lamda-7).1+(4lamda-7).4+(7-9lamda).(-9)=0`
`implies lamda-7+16lamda-28+81lamda-63=0`
`implies 98lamda=98implieslamda=1`
So the coordinates of `L` are `(-4,1,-3)`
`:.` Required distance `=sqrt((-4-2)^(2)+(1-4)^(2)+(-3+1)^(2))`
`=sqrt(36+9+4)=7`units