以原点 $O$ 为参考点，点 $A$、$B$ 和 $C$ 的位置向量分别表示为 $\overrightarrow{OA}$、$\overrightarrow{OB}$ 和 $\overrightarrow{OC}$，使得
$\overrightarrow{OA} = (1 - \lambda)\overrightarrow{OB} + \lambda\overrightarrow{OC},$
其中 $\lambda$ 是一个非零常数。

1. Find an expression for the displacement vector $\overrightarrow{AB}$ in terms of $\overrightarrow{OB}$ and $\overrightarrow{OC}$.

$\begin{aligned} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \overrightarrow{OB} - [(1 - \lambda)\overrightarrow{OB} + \lambda\overrightarrow{OC}] \\ &= \overrightarrow{OB} - \overrightarrow{OB} + \lambda\overrightarrow{OB} - \lambda\overrightarrow{OC} \\ &= \lambda\overrightarrow{OB} - \lambda\overrightarrow{OC} \end{aligned}$

2. Show that $\overrightarrow{AB}$ is a scalar multiple of $\overrightarrow{CB}$ and state the geometric conclusion.

$\begin{aligned} \overrightarrow{AB} &= \lambda(\overrightarrow{OB} - \overrightarrow{OC}) \\ &= \lambda\overrightarrow{CB} \end{aligned}$
Since $\lambda$ is a non-zero constant, $\overrightarrow{AB}$ is parallel to $\overrightarrow{CB}$.
Because point $B$ is common to both vectors, points $A$, $B$, and $C$ are collinear.

1. Find an expression for $\overrightarrow{OC}$ in terms of $\overrightarrow{OA}$ and $\overrightarrow{OB}$.

Given $\overrightarrow{OA} = (1 - \lambda)\overrightarrow{OB} + \lambda\overrightarrow{OC}$ and $\lambda = \frac{1}{6}$,
$\overrightarrow{OA} = \frac{5}{6}\overrightarrow{OB} + \frac{1}{6}\overrightarrow{OC}$
$\overrightarrow{OC} = 6\overrightarrow{OA} - 5\overrightarrow{OB}$

2. Apply the Ratio Theorem to express $\overrightarrow{OD}$ in terms of $\overrightarrow{OA}$ and $\overrightarrow{OB}$.

Since $D$ divides $AC$ internally in the ratio $AD : DC = 1 : 4$, by the Ratio Theorem:
$\overrightarrow{OD} = \frac{4\overrightarrow{OA} + 1\overrightarrow{OC}}{1 + 4}$
Substituting the expression for $\overrightarrow{OC}$ into the equation:
$\overrightarrow{OD} = \frac{4\overrightarrow{OA} + (6\overrightarrow{OA} - 5\overrightarrow{OB})}{5}$
$\overrightarrow{OD} = \frac{10\overrightarrow{OA} - 5\overrightarrow{OB}}{5} = 2\overrightarrow{OA} - \overrightarrow{OB}$

3. Substitute the column vectors to find the position vector of $D$.

$\overrightarrow{OD} = 2 \begin{pmatrix} 0 \\ -2 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}$
$\overrightarrow{OD} = \begin{pmatrix} 0 \\ -4 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} -1 \\ -5 \\ 2 \end{pmatrix}$

1. Find the displacement vectors $\overrightarrow{AB}$ and $\overrightarrow{AE}$.

Using the position vectors from part (ii) and the coordinates of $E$:
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} - \begin{pmatrix} 0 \\ -2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}$
$\overrightarrow{AE} = \overrightarrow{OE} - \overrightarrow{OA} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 0 \\ -2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$

2. Calculate the cross product of $\overrightarrow{AB}$ and $\overrightarrow{AE}$.

$\overrightarrow{AB} \times \overrightarrow{AE} = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} (3)(2) - (-2)(1) \\ (-2)(1) - (1)(2) \\ (1)(1) - (3)(1) \end{pmatrix} = \begin{pmatrix} 8 \\ -4 \\ -2 \end{pmatrix}$

3. Compute the exact area of the triangle using the magnitude of the cross product.

$\text{Area of } \triangle ABE = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AE}|$
$= \frac{1}{2} \sqrt{8^2 + (-4)^2 + (-2)^2}$
$= \frac{1}{2} \sqrt{64 + 16 + 4} = \frac{1}{2} \sqrt{84}$
$= \frac{1}{2} (2\sqrt{21}) = \sqrt{21}$

1. Find the displacement vector $\overrightarrow{AB}$.

Using the position vectors given in part (ii):
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} - \begin{pmatrix} 0 \\ -2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}$

2. Calculate the length of projection of $\overrightarrow{OE}$ onto $\overrightarrow{AB}$.

$\text{Length of projection} = \frac{|\overrightarrow{OE} \cdot \overrightarrow{AB}|}{|\overrightarrow{AB}|}$

$= \frac{\left| \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} \right|}{\sqrt{1^2 + 3^2 + (-2)^2}}$

$= \frac{|(1)(1) + (-1)(3) + (2)(-2)|}{\sqrt{1 + 9 + 4}}$

$= \frac{|1 - 3 - 4|}{\sqrt{14}}$

$= \frac{6}{\sqrt{14}} = \frac{6\sqrt{14}}{14} = \frac{3\sqrt{14}}{7}$