以原点 $O$ 为基准，点 $A$ 和点 $B$ 的位置向量分别为 $\mathbf{a}$ 和 $\mathbf{b}$，使得夹角 $\angle AOB = 30^\circ$ 且 $|\mathbf{a}| = 4|\mathbf{b}|$。

1. Express the displacement vectors $\vec{AB}$ and $\vec{AC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

$\vec{AB} = \mathbf{b} - \mathbf{a}$
$\vec{AC} = \mathbf{c} - \mathbf{a} = m\mathbf{a} + n\mathbf{b} - \mathbf{a} = (m-1)\mathbf{a} + n\mathbf{b}$

2. State the formula for the area of triangle $ABC$ using the vector product.

$\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|$
$\vec{AB} \times \vec{AC} = (\mathbf{b} - \mathbf{a}) \times ((m-1)\mathbf{a} + n\mathbf{b})$

3. Expand and simplify the vector product using its algebraic properties.

$\vec{AB} \times \vec{AC} = (m-1)(\mathbf{b} \times \mathbf{a}) + n(\mathbf{b} \times \mathbf{b}) - (m-1)(\mathbf{a} \times \mathbf{a}) - n(\mathbf{a} \times \mathbf{b})$
Using $\mathbf{b} \times \mathbf{b} = \mathbf{0}$, $\mathbf{a} \times \mathbf{a} = \mathbf{0}$, and $\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})$:
$\vec{AB} \times \vec{AC} = -(m-1)(\mathbf{a} \times \mathbf{b}) - n(\mathbf{a} \times \mathbf{b}) = -(m+n-1)(\mathbf{a} \times \mathbf{b})$

4. Calculate the magnitude of the vector product $\mathbf{a} \times \mathbf{b}$.

$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin 30^\circ$
Substitute $|\mathbf{a}| = 4|\mathbf{b}|$ and $\sin 30^\circ = \frac{1}{2}$:
$|\mathbf{a} \times \mathbf{b}| = (4|\mathbf{b}|)|\mathbf{b}|\left(\frac{1}{2}\right) = 2|\mathbf{b}|^2$

5. Compute the final area, resolving the absolute value using the given conditions on $m$ and $n$.

$\text{Area} = \frac{1}{2} |-(m+n-1)(\mathbf{a} \times \mathbf{b})| = \frac{1}{2} (m+n-1) |\mathbf{a} \times \mathbf{b}|$
(since $m \ge 1$ and $n \ge 1$ are positive integers, $m+n-1 > 0$)
$\text{Area} = \frac{1}{2} (m+n-1)(2|\mathbf{b}|^2) = (m+n-1)|\mathbf{b}|^2$

1. Find the position vector of $D$.

Since $D$ is the mid-point of $OA$,
$\overrightarrow{OD} = \frac{1}{2}\overrightarrow{OA} = \frac{1}{2}\mathbf{a}$

2. Find the position vector of $E$.

Since $E$ lies on $OB$ such that $OE : EB = 3 : 2$,
$\overrightarrow{OE} = \frac{3}{3 + 2}\overrightarrow{OB} = \frac{3}{5}\mathbf{b}$

1. Show the vector equation of the line $BD$.

From part (ii), $\overrightarrow{OD} = \frac{1}{2}\mathbf{a}$.
The line $BD$ passes through $B(\mathbf{b})$ and $D\left(\frac{1}{2}\mathbf{a}\right)$. Its vector equation with a parameter $s$ is:
$\mathbf{r} = \overrightarrow{OB} + s\overrightarrow{BD} = \mathbf{b} + s\left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right)$
$\mathbf{r} = \frac{s}{2}\mathbf{a} + (1-s)\mathbf{b}$
Let $\lambda = \frac{s}{2}$, which means $s = 2\lambda$:
$\mathbf{r} = \lambda\mathbf{a} + (1-2\lambda)\mathbf{b} \quad \text{(shown)}$

2. Find the vector equation of the line $AE$.

From part (ii), $\overrightarrow{OE} = \frac{3}{5}\mathbf{b}$.
The line $AE$ passes through $A(\mathbf{a})$ and $E\left(\frac{3}{5}\mathbf{b}\right)$. Its vector equation with parameter $\mu$ is:
$\mathbf{r} = \overrightarrow{OA} + \mu\overrightarrow{AE} = \mathbf{a} + \mu\left(\frac{3}{5}\mathbf{b} - \mathbf{a}\right)$
$\mathbf{r} = (1-\mu)\mathbf{a} + \frac{3}{5}\mu\mathbf{b}$

3. Equate the two line equations to locate the intersection point $F$.

At the intersection point $F$, the position vectors from both lines must be equal:
$\lambda\mathbf{a} + (1-2\lambda)\mathbf{b} = (1-\mu)\mathbf{a} + \frac{3}{5}\mu\mathbf{b}$
Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel vectors (the angle between them is $30^\circ$), we can equate their respective coefficients:
$\lambda = 1-\mu \quad \text{--- (1)}$
$1-2\lambda = \frac{3}{5}\mu \quad \text{--- (2)}$

4. Solve for the parameters and state the position vector of $F$.

Substitute (1) into (2):
$1 - 2(1-\mu) = \frac{3}{5}\mu$
$1 - 2 + 2\mu = \frac{3}{5}\mu$
$-1 + 2\mu = \frac{3}{5}\mu$
$\frac{7}{5}\mu = 1 \implies \mu = \frac{5}{7}$
Substitute $\mu = \frac{5}{7}$ back into the equation for line $AE$ to find $F$:
$\overrightarrow{OF} = \left(1 - \frac{5}{7}\right)\mathbf{a} + \frac{3}{5}\left(\frac{5}{7}\right)\mathbf{b}$
$\overrightarrow{OF} = \frac{2}{7}\mathbf{a} + \frac{3}{7}\mathbf{b}$