相对于原点 $O$，点 $A$、$B$ 和 $C$ 的位置向量分别为 $\mathbf{a}$、$\mathbf{b}$ 和 $\mathbf{c}$，其中 $\mathbf{a}$、$\mathbf{b}$ 和 $\mathbf{c}$ 是互不平行的非零向量。点 $A$、$B$ 和 $C$ 不共线。 三等分点是将线段内分，其比例为 $1:2$ 或 $2:1$ 的点。
假设另外两个点 $D$ 和 $E$ 分别是线段 $AB$ 和 $AC$ 的三等分点，且两个点分别比靠近 $B$ 和 $C$ 更靠近 $A$。直线 $BE$ 与 $CD$ 相交于点 $F$。

1. Determine the position vectors of the points of trisection $D$ and $E$ using the ratio theorem.

Since $D$ divides $AB$ internally in the ratio $1:2$ (nearer to $A$):
$\mathbf{d} = \frac{2\mathbf{a} + 1\mathbf{b}}{1 + 2} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$

Similarly, since $E$ divides $AC$ internally in the ratio $1:2$:
$\mathbf{e} = \frac{2\mathbf{a} + 1\mathbf{c}}{1 + 2} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{c}$

2. Formulate the vector equation of the line $BE$.

The line passes through $B$ (position vector $\mathbf{b}$) and $E$ (position vector $\mathbf{e}$). Using the parameter $\lambda$, the vector equation is:
$\mathbf{r} = (1-\lambda)\mathbf{b} + \lambda\mathbf{e}$
Substituting $\mathbf{e}$:
$\mathbf{r} = (1-\lambda)\mathbf{b} + \lambda\left(\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{c}\right)$
$\mathbf{r} = \frac{2}{3}\lambda\mathbf{a} + (1-\lambda)\mathbf{b} + \frac{1}{3}\lambda\mathbf{c} \quad \text{(Shown)}$

3. Formulate the vector equation of the line $CD$.

The line passes through $C$ (position vector $\mathbf{c}$) and $D$ (position vector $\mathbf{d}$). Using the parameter $\mu$, the vector equation is:
$\mathbf{r} = (1-\mu)\mathbf{c} + \mu\mathbf{d}$
Substituting $\mathbf{d}$:
$\mathbf{r} = (1-\mu)\mathbf{c} + \mu\left(\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}\right)$
$\mathbf{r} = \frac{2}{3}\mu\mathbf{a} + \frac{1}{3}\mu\mathbf{b} + (1-\mu)\mathbf{c} \quad \text{(Shown)}$

4. Equate the vector equations at the intersection point $F$ to find $\alpha$, $\beta$, and $\gamma$.

At the point of intersection $F$, the position vectors from both line equations must be equal:
$\frac{2}{3}\lambda\mathbf{a} + (1-\lambda)\mathbf{b} + \frac{1}{3}\lambda\mathbf{c} = \frac{2}{3}\mu\mathbf{a} + \frac{1}{3}\mu\mathbf{b} + (1-\mu)\mathbf{c}$

Rearranging to group the vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$:
$\left(\frac{2}{3}\lambda - \frac{2}{3}\mu\right)\mathbf{a} + \left(1 - \lambda - \frac{1}{3}\mu\right)\mathbf{b} + \left(\frac{1}{3}\lambda - (1 - \mu)\right)\mathbf{c} = \mathbf{0}$

Comparing this to the given expression $\alpha\mathbf{a} + \beta\mathbf{b} + \gamma\mathbf{c} = \mathbf{0}$:
$\alpha = \frac{2}{3}\lambda - \frac{2}{3}\mu$
$\beta = 1 - \lambda - \frac{1}{3}\mu$
$\gamma = \frac{1}{3}\lambda + \mu - 1$

1. State the vector relationship for parallelogram $OACB$.

Since the vertices of parallelogram $OACB$ are given in cyclical order, the diagonals $OC$ and $AB$ bisect each other, and opposite sides are equal in magnitude and direction. Thus, $\vec{OA} + \vec{OB} = \vec{OC}$.
$\mathbf{c} = \mathbf{a} + \mathbf{b}$

2. Substitute $\mathbf{c}$ into the vector equations of the lines to express them strictly in terms of $\mathbf{a}$ and $\mathbf{b}$.

From part (i), substitute $\mathbf{c} = \mathbf{a} + \mathbf{b}$ into the line equations:
For line $BE$:
$\mathbf{r} = \frac{2}{3}\lambda\mathbf{a} + (1-\lambda)\mathbf{b} + \frac{1}{3}\lambda(\mathbf{a} + \mathbf{b})$
$\mathbf{r} = \lambda\mathbf{a} + \left(1 - \frac{2}{3}\lambda\right)\mathbf{b}$
For line $CD$:
$\mathbf{r} = \frac{2}{3}\mu\mathbf{a} + \frac{1}{3}\mu\mathbf{b} + (1-\mu)(\mathbf{a} + \mathbf{b})$
$\mathbf{r} = \left(1 - \frac{1}{3}\mu\right)\mathbf{a} + \left(1 - \frac{2}{3}\mu\right)\mathbf{b}$

3. Equate the coefficients to solve for the parameter and deduce the position vector of $F$.

Since $F$ is the point of intersection and vectors $\mathbf{a}$ and $\mathbf{b}$ are non-zero and not parallel, we can equate their coefficients:
Comparing coefficients of $\mathbf{b}$:
$1 - \frac{2}{3}\lambda = 1 - \frac{2}{3}\mu \implies \lambda = \mu$
Comparing coefficients of $\mathbf{a}$:
$\lambda = 1 - \frac{1}{3}\mu$
Substitute $\lambda = \mu$:
$\lambda = 1 - \frac{1}{3}\lambda \implies \frac{4}{3}\lambda = 1 \implies \lambda = \frac{3}{4}$
Substitute $\lambda = \frac{3}{4}$ into the simplified equation for line $BE$:
$\mathbf{r}_F = \frac{3}{4}\mathbf{a} + \left(1 - \frac{2}{3}\left(\frac{3}{4}\right)\right)\mathbf{b} = \frac{3}{4}\mathbf{a} + \frac{1}{2}\mathbf{b}$