以原点 $O$ 为基准，点 $A$ 和 $B$ 满足 $\overrightarrow{OA} = \mathbf{a}$ 且 $\overrightarrow{OB} = \mathbf{b}$。点 $C$ 在 $OB$ 上，满足 $\overrightarrow{OC} = p\overrightarrow{OB}$，其中 $p$ 为常数。$D$ 在 $AC$ 上，满足 $AD : DC = 2 : 3$，且 $E$ 在 $AB$ 上，满足 $AE : EB = 1 : 3$。

1. Find $\overrightarrow{OD}$ using the given ratio $AD : DC = 2 : 3$.

Given $\overrightarrow{OC} = p\overrightarrow{OB} = p\mathbf{b}$.
Using the ratio theorem on line segment $AC$:
$\overrightarrow{OD} = \frac{3\overrightarrow{OA} + 2\overrightarrow{OC}}{2 + 3}$
$\overrightarrow{OD} = \frac{3\mathbf{a} + 2(p\mathbf{b})}{5} = \frac{3}{5}\mathbf{a} + \frac{2p}{5}\mathbf{b}$

2. Find $\overrightarrow{OE}$ using the given ratio $AE : EB = 1 : 3$.

Using the ratio theorem on line segment $AB$:
$\overrightarrow{OE} = \frac{3\overrightarrow{OA} + 1\overrightarrow{OB}}{1 + 3}$
$\overrightarrow{OE} = \frac{3\mathbf{a} + \mathbf{b}}{4} = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$

1. Concisely find $\overrightarrow{OD}$ and $\overrightarrow{OE}$ using the ratio theorem (as required from part (i)).

Given $AD : DC = 2 : 3$ and $\overrightarrow{OC} = p\mathbf{b}$:
$\overrightarrow{OD} = \frac{3\overrightarrow{OA} + 2\overrightarrow{OC}}{2+3} = \frac{3}{5}\mathbf{a} + \frac{2p}{5}\mathbf{b}$

Given $AE : EB = 1 : 3$:
$\overrightarrow{OE} = \frac{3\overrightarrow{OA} + 1\overrightarrow{OB}}{1+3} = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$

2. Apply the collinearity condition for points $O$, $D$, and $E$.

Since $O$, $D$, and $E$ are collinear, there exists a scalar $k$ such that $\overrightarrow{OD} = k\overrightarrow{OE}$.
$\frac{3}{5}\mathbf{a} + \frac{2p}{5}\mathbf{b} = k\left(\frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\right)$

3. Compare the coefficients of $\mathbf{a}$ and $\mathbf{b}$ to solve for $p$.

Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel vectors, we can compare their coefficients.
Comparing coefficients of $\mathbf{a}$:
$\frac{3}{5} = \frac{3}{4}k \implies k = \frac{4}{5}$

Comparing coefficients of $\mathbf{b}$:
$\frac{2p}{5} = \frac{1}{4}k$
Substitute $k = \frac{4}{5}$ into the equation:
$\frac{2p}{5} = \frac{1}{4}\left(\frac{4}{5}\right)$
$\frac{2p}{5} = \frac{1}{5}$
$2p = 1 \implies p = \frac{1}{2}$

1. Use the ratio theorem to find the position vector of $E$.

Since $AE : EB = 1 : 3$, we apply the ratio theorem to find $\overrightarrow{OE}$:
$\overrightarrow{OE} = \frac{3 \overrightarrow{OA} + 1 \overrightarrow{OB}}{1 + 3} = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$

2. Set up the formula for the shortest distance from a point to a line using the cross product.

The shortest distance $d$ from a point $E$ to the line passing through $O$ and $B$ is given by the perpendicular distance formula:
$d = \frac{|\overrightarrow{OE} \times \overrightarrow{OB}|}{|\overrightarrow{OB}|}$

3. Evaluate the cross product and determine the value of the constant $k$.

Substitute $\overrightarrow{OE}$ and $\overrightarrow{OB} = \mathbf{b}$ into the cross product:
$\overrightarrow{OE} \times \overrightarrow{OB} = \left( \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b} \right) \times \mathbf{b}$
Using the distributive property and the fact that $\mathbf{b} \times \mathbf{b} = \mathbf{0}$:
$\overrightarrow{OE} \times \overrightarrow{OB} = \frac{3}{4}(\mathbf{a} \times \mathbf{b}) + \frac{1}{4}(\mathbf{b} \times \mathbf{b}) = \frac{3}{4}(\mathbf{a} \times \mathbf{b})$
Given that the length $|\overrightarrow{OB}| = OB = 5$, substitute this into the distance formula:
$d = \frac{\left| \frac{3}{4}(\mathbf{a} \times \mathbf{b}) \right|}{5} = \frac{3}{20}|\mathbf{a} \times \mathbf{b}|$
Comparing this to $k|\mathbf{a} \times \mathbf{b}|$, we obtain $k = \frac{3}{20}$.

1. State the geometrical interpretation

$|\mathbf{a} \cdot \hat{\mathbf{b}}|$ represents the length of the projection of the vector $\mathbf{a}$ (which is $\overrightarrow{OA}$) onto the direction of $\mathbf{b}$ (which is the line $OB$).