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 $AC$.

Let $O$ be the origin of reference.
Then, $\overrightarrow {OA} = \hat i - 2\hat j - 8\hat k,\overrightarrow {OB} = 5\hat i - 2\hat k$ and $\overrightarrow {OC} = 11\hat i + 3\hat j + 7\hat k$

$\therefore$ $\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = (5\hat i - 2\hat k) - (\hat i - 2\widehat j - 8\hat k) = 4\hat i + 2\widehat j + 6\hat k$

$\overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} = (11\hat i + 3\hat j + 7\hat k) - (\hat i - 2\widehat j - 8\hat k) = 10\hat i + 5\widehat j + 15\hat k$

$\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{llllllllllllllllllll}{\hat i}&{\hat j}&{\hat k}\\4&2&6\\{10}&5&{15}\end{array}} \right|$
$= (30 - 30)\hat i - (60 - 60)\hat j + (20 - 20)\hat k = 0$

$\Rightarrow \overrightarrow {AC}$ and $\overrightarrow {AB}$

are parallel and having $A$ as the common point. Hence, the points $A,B,C$ are collinear.

Let $B$ divides $[AC]$ in the ratio $\lambda :1$.

$\Rightarrow \cfrac{{\lambda ({\rm{position vector of }}C) + 1 \times ({\rm{position vector of }}A)}}{{\lambda + 1}}$

$= \cfrac{1}{{\lambda + 1}}\left\{ {\lambda (11\hat i + 3\hat j + 7\hat k) + 1(\hat i - 2\hat j - 8\hat k)} \right\}$

$= \left( {\cfrac{{11\lambda + 1}}{{\lambda + 1}}} \right)\hat i + \left( {\cfrac{{3\lambda - 2}}{{\lambda + 1}}} \right)\hat j + \left( {\cfrac{{7\lambda - 8}}{{\lambda + 1}}} \right)\hat k = 5\hat i + 0\widehat j - 2\hat k$

$\Rightarrow \cfrac{{11\lambda + 1}}{{\lambda + 1}} = 5,\cfrac{{3\lambda - 2}}{{\lambda + 1}} = 0,\cfrac{{7\lambda - 8}}{{\lambda + 1}} = - 2$.

$\Rightarrow 11\lambda + I = 5\lambda + 5,3\lambda = 2,7\lambda - 8 = - 2\lambda - 2$

$\Rightarrow 6\lambda = 4,\lambda = \cfrac{2}{3},9\lambda = 6 \Rightarrow \lambda = \cfrac{2}{3} = 2:3$