Find the length of the perpendicular drawn from point `(2,3,4)` to line `(4-x)/2=y/6=(1-z)/3dot`

Explanation

Let P be the foot of the perpendicular from `A(2, 3, 4)` to the given line `l` whose equation is
` " "(4-x)/(2)=(y)/(6)=(1-z)/(3) or (x-4)/(-2)=(y)/(6)=(z-1)/(-3)=k" (say)"" "`(i)
`therefore" "x=4-2k, y=6k, z=1-3k`
As P lies on (i), coordinates of P are (4 -2k, 6k, 1-3k) for some value of k.
The direction ratios fo AP are
`" "(4-2k-2, 6k-3, 1-3k-4)or (2-2k, 6k-3, -3-3k)`
Also, the direction ratios of `l` are -2, 6 and -3.
Since `APbot l`, we have
`" "-2(2-2k)+6(6k-3)-3(-3-3k)=0`
or `" "-4+4k+36k-18+9+9k=0 or 49k-13=0 or k=13//49`
We have `" "AP^(2)=(4-2k-2)^(2)+(6k-3)^(2)+(1-3k-4)^(2)`
`" "=(2-2k)^(2)+(6k-3)^(2)+(-3-3k)^(2)`
`" "=4-8k+4k^(2)+36k^(2)-36k+9+9+18k+9k^(2)`
`" "=22-26k+49k^(2)`
`" "=22-26((13)/(49))+49((13)/(49))^(2)`
`" "=(22xx49-26xx13+13^(2))/(49)=(909)/(49)`
`" "AP=(3)/(7)sqrt(101)`