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

The given line is $x+5 = \dfrac{y+3}{4} = \dfrac{z-6}{-9} = \lambda$.

Any point on the line is $P(\lambda - 5, 4\lambda - 3, -9\lambda + 6)$.

Let $A = (2, 4, -1)$. The direction vector of the line is $\vec{v} = (1, 4, -9)$.

The vector $\vec{AP} = (\lambda - 5 - 2, 4\lambda - 3 - 4, -9\lambda + 6 - (-1)) = (\lambda - 7, 4\lambda - 7, -9\lambda + 7)$.

Since $\vec{AP}$ is perpendicular to the line, $\vec{AP} \cdot \vec{v} = 0$.

$1(\lambda - 7) + 4(4\lambda - 7) - 9(-9\lambda + 7) = 0$.

$\lambda - 7 + 16\lambda - 28 + 81\lambda - 63 = 0$.

$98\lambda - 98 = 0 \Rightarrow \lambda = 1$.

Substituting $\lambda = 1$ into the coordinates of $P$:

$x = 1 - 5 = -4$, $y = 4(1) - 3 = 1$, $z = -9(1) + 6 = -3$.

The foot of the perpendicular is $(-4, 1, -3)$.

Answer: $(-4,1,-3)$