Show that the points $A (1,-2,-8), B (5,0,-2)$ and $C (11,3,7)$ are collinear,and find the ratio in which $B$ divides $AC$.

(2:3) The given points are $A (1,-2,-8), B (5,0,-2)$ and $C (11,3,7)$.
$\overrightarrow{AB} = (5-1)\hat{i} + (0+2)\hat{j} + (-2+8)\hat{k} = 4\hat{i} + 2\hat{j} + 6\hat{k}$.
$\overrightarrow{BC} = (11-5)\hat{i} + (3-0)\hat{j} + (7+2)\hat{k} = 6\hat{i} + 3\hat{j} + 9\hat{k}$.
$\overrightarrow{AC} = (11-1)\hat{i} + (3+2)\hat{j} + (7+8)\hat{k} = 10\hat{i} + 5\hat{j} + 15\hat{k}$.
$|\overrightarrow{AB}| = \sqrt{4^2 + 2^2 + 6^2} = \sqrt{16+4+36} = \sqrt{56} = 2\sqrt{14}$.
$|\overrightarrow{BC}| = \sqrt{6^2 + 3^2 + 9^2} = \sqrt{36+9+81} = \sqrt{126} = 3\sqrt{14}$.
$|\overrightarrow{AC}| = \sqrt{10^2 + 5^2 + 15^2} = \sqrt{100+25+225} = \sqrt{350} = 5\sqrt{14}$.
Since $|\overrightarrow{AC}| = |\overrightarrow{AB}| + |\overrightarrow{BC}|$,the points $A, B,$ and $C$ are collinear.
Let point $B$ divide $AC$ in the ratio $\lambda : 1$. Using the section formula:
$\vec{b} = \frac{\lambda\vec{c} + 1\vec{a}}{\lambda + 1}$.
$5\hat{i} + 0\hat{j} - 2\hat{k} = \frac{\lambda(11\hat{i} + 3\hat{j} + 7\hat{k}) + (1\hat{i} - 2\hat{j} - 8\hat{k})}{\lambda + 1}$.
Equating the $\hat{j}$ components:
$0 = \frac{3\lambda - 2}{\lambda + 1} \Rightarrow 3\lambda - 2 = 0 \Rightarrow \lambda = \frac{2}{3}$.
Thus,the ratio is $2:3$.