The foot of perpendicular from point (2, 4, -1) on the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$ is:

(-4, 1, -3)

$\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=λ$  DR = (1, 4, -9)

$x=λ-5,y=4λ-3,z=-9λ+6$

Q ⇒ (a, b, c) = $(λ−5,4λ−3,-9λ+6)$ ...(i)

DR's of PQ = $(λ−5-2,4λ−3-4,-9λ+6+1)$

$=(λ−7,4λ−7,-9λ+7)$

(DR's of PQ).(DR's of line) = 0

$((λ−7).1+(4λ−7).4-9((-9λ)+7))=0$

$λ−7+16λ-28+81λ-63=0$

$98λ-98=0⇒98λ=98⇒λ=\frac{98}{98}=1$ ...(ii)

Putting eq. (ii) in (i)

$((1)−5,4(1)−3,-9(1)+6)$

$⇒(-4, 1, -3)$