Find the distance of a point $(2, 4, -1)$ from the line $\frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9}$.

(C) Let the equation of the line be $\frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9} = \lambda$.
Then,any point $L$ on the line is given by $x = \lambda - 5$,$y = 4\lambda - 3$,$z = 6 - 9\lambda$.
Let $P$ be the point $(2, 4, -1)$. The direction ratios of the line segment $PL$ are $(\lambda - 5 - 2, 4\lambda - 3 - 4, 6 - 9\lambda - (-1))$,which simplifies to $(\lambda - 7, 4\lambda - 7, 7 - 9\lambda)$.
Since $PL$ is perpendicular to the given line,the dot product of the direction ratios of $PL$ and the line $(1, 4, -9)$ must be zero:
$1(\lambda - 7) + 4(4\lambda - 7) - 9(7 - 9\lambda) = 0$.
Expanding this,we get $\lambda - 7 + 16\lambda - 28 - 63 + 81\lambda = 0$.
Combining like terms,$98\lambda - 98 = 0$,which gives $\lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates of $L$,we get $L = (1 - 5, 4(1) - 3, 6 - 9(1)) = (-4, 1, -3)$.
The distance $PL$ is $\sqrt{(-4 - 2)^2 + (1 - 4)^2 + (-3 - (-1))^2} = \sqrt{(-6)^2 + (-3)^2 + (-2)^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$ units.