Find the foot of the perpendicular from the point $(2, 3, -8)$ to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$. Also,find the perpendicular distance from the given point to the line.

(N/A) We have the equation of the line as $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$.
Rewriting it in standard form: $\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}=\lambda$.
Thus,any point on the line is given by $x = -2\lambda + 4$,$y = 6\lambda$,and $z = -3\lambda + 1$.
Let $L$ be the foot of the perpendicular from $P(2, 3, -8)$ to the line. So,the coordinates of $L$ are $(4-2\lambda, 6\lambda, 1-3\lambda)$.
The direction ratios of the line $PL$ are $(4-2\lambda-2, 6\lambda-3, 1-3\lambda+8)$,which simplifies to $(2-2\lambda, 6\lambda-3, 9-3\lambda)$.
The direction ratios of the given line are $(-2, 6, -3)$.
Since $PL$ is perpendicular to the line,the dot product of their direction ratios is zero:
$-2(2-2\lambda) + 6(6\lambda-3) - 3(9-3\lambda) = 0$.
$-4 + 4\lambda + 36\lambda - 18 - 27 + 9\lambda = 0$.
$49\lambda - 49 = 0 \Rightarrow \lambda = 1$.
Substituting $\lambda = 1$ into the coordinates of $L$,we get $L = (4-2(1), 6(1), 1-3(1)) = (2, 6, -2)$.
The perpendicular distance $PL$ is the distance between $P(2, 3, -8)$ and $L(2, 6, -2)$:
$PL = \sqrt{(2-2)^2 + (6-3)^2 + (-2 - (-8))^2} = \sqrt{0^2 + 3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$ units.