Find the coordinates of the foot of the perpendicular drawn from the point $A(1, 8, 4)$ to the line joining the points $B(0, -1, 3)$ and $C(2, -3, -1)$.

(N/A) Let $L$ be the foot of the perpendicular drawn from point $A(1, 8, 4)$ to the line passing through $B(0, -1, 3)$ and $C(2, -3, -1)$.
The direction ratios of the line $BC$ are $(2-0, -3-(-1), -1-3)$,which are $(2, -2, -4)$.
The equation of the line $BC$ passing through $B(0, -1, 3)$ with direction ratios $(2, -2, -4)$ is given by:
$\frac{x-0}{2} = \frac{y+1}{-2} = \frac{z-3}{-4} = \lambda$
Any point $L$ on the line $BC$ can be represented as:
$L = (2\lambda, -2\lambda-1, -4\lambda+3)$
The direction ratios of the line $AL$ are:
$(2\lambda-1, -2\lambda-1-8, -4\lambda+3-4) = (2\lambda-1, -2\lambda-9, -4\lambda-1)$
Since $AL \perp BC$,the dot product of their direction ratios is zero:
$2(2\lambda-1) - 2(-2\lambda-9) - 4(-4\lambda-1) = 0$
$4\lambda - 2 + 4\lambda + 18 + 16\lambda + 4 = 0$
$24\lambda + 20 = 0$
$24\lambda = -20$
$\lambda = -\frac{20}{24} = -\frac{5}{6}$
Substituting $\lambda = -\frac{5}{6}$ into the coordinates of $L$:
$x = 2(-\frac{5}{6}) = -\frac{5}{3}$
$y = -2(-\frac{5}{6}) - 1 = \frac{5}{3} - 1 = \frac{2}{3}$
$z = -4(-\frac{5}{6}) + 3 = \frac{10}{3} + 3 = \frac{19}{3}$
Thus,the coordinates of the foot of the perpendicular are $(-\frac{5}{3}, \frac{2}{3}, \frac{19}{3})$.