设 $O$ 为原点，点 $A$ 和 $B$ 满足 $\overrightarrow{OA} = \mathbf{a}$ 且 $\overrightarrow{OB} = \mathbf{b}$。$OA$ 上的点 $C$ 满足 $OC : OA = 1 : 3$。直线 $l$ 经过点 $A$ 和 $B$。已知角 $BOA = 60^\circ$ 且 $|\mathbf{a}| = 3|\mathbf{b}|$。

1. Expand the scalar product $(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{b} - \mathbf{a})$ and apply the dot product formula.

$(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{b} - \mathbf{a}) = \mathbf{b} \cdot \mathbf{b} - 2(\mathbf{a} \cdot \mathbf{b}) + \mathbf{a} \cdot \mathbf{a}$
$= |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos 60^\circ + |\mathbf{a}|^2$

2. Substitute $|\mathbf{a}| = 3|\mathbf{b}|$ and evaluate the expression in terms of $|\mathbf{b}|$.

$= |\mathbf{b}|^2 - 2(3|\mathbf{b}|)|\mathbf{b}|\left(\frac{1}{2}\right) + (3|\mathbf{b}|)^2$
$= |\mathbf{b}|^2 - 3|\mathbf{b}|^2 + 9|\mathbf{b}|^2$
$= 7|\mathbf{b}|^2$

3. Relate the result to $|\mathbf{b} - \mathbf{a}|^2$ and take the square root to express $|\mathbf{b} - \mathbf{a}|$ in the required form.

Since $(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{b} - \mathbf{a}) = |\mathbf{b} - \mathbf{a}|^2$:
$|\mathbf{b} - \mathbf{a}|^2 = 7|\mathbf{b}|^2$
$|\mathbf{b} - \mathbf{a}| = \sqrt{7}|\mathbf{b}|$
$\therefore k = \sqrt{7}$

1. Concisely determine $|\mathbf{b} - \mathbf{a}|$ (the result from part i) as it is required for the distance formula.

Using the dot product expansion,
$|\mathbf{b} - \mathbf{a}|^2 = (\mathbf{b} - \mathbf{a}) \cdot (\mathbf{b} - \mathbf{a}) = |\mathbf{b}|^2 - 2\mathbf{a} \cdot \mathbf{b} + |\mathbf{a}|^2$
Given angle $BOA = 60^\circ$ and $|\mathbf{a}| = 3|\mathbf{b}|$:
$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos 60^\circ = (3|\mathbf{b}|)|\mathbf{b}|(0.5) = 1.5|\mathbf{b}|^2$
$|\mathbf{b} - \mathbf{a}|^2 = |\mathbf{b}|^2 - 2(1.5|\mathbf{b}|^2) + (3|\mathbf{b}|)^2 = 7|\mathbf{b}|^2 \implies |\mathbf{b} - \mathbf{a}| = \sqrt{7}|\mathbf{b}|$

2. Find the displacement vector $\overrightarrow{CA}$.

Since $C$ lies on $OA$ such that $OC : OA = 1 : 3$, $\overrightarrow{OC} = \frac{1}{3}\mathbf{a}$.
Thus, $\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = \mathbf{a} - \frac{1}{3}\mathbf{a} = \frac{2}{3}\mathbf{a}$.

3. Set up the formula for the shortest distance from $C$ to line $l$.

The line $l$ passes through $A$ and $B$, so its direction vector is $\mathbf{d} = \mathbf{b} - \mathbf{a}$.
The shortest distance $d$ from point $C$ to line $l$ is given by
$d = \frac{|\overrightarrow{CA} \times \mathbf{d}|}{|\mathbf{d}|} = \frac{\left|\frac{2}{3}\mathbf{a} \times (\mathbf{b} - \mathbf{a})\right|}{|\mathbf{b} - \mathbf{a}|}$

4. Expand and simplify the cross product in the numerator.

Using the distributive property of the cross product:
$\overrightarrow{CA} \times \mathbf{d} = \frac{2}{3}\mathbf{a} \times (\mathbf{b} - \mathbf{a}) = \frac{2}{3}(\mathbf{a} \times \mathbf{b}) - \frac{2}{3}(\mathbf{a} \times \mathbf{a})$
Since $\mathbf{a} \times \mathbf{a} = \mathbf{0}$, this simplifies to $\frac{2}{3}(\mathbf{a} \times \mathbf{b})$.

5. Evaluate the magnitude of the cross product.

Using the definition of the cross product magnitude:
$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin 60^\circ = (3|\mathbf{b}|)|\mathbf{b}|\left(\frac{\sqrt{3}}{2}\right) = \frac{3\sqrt{3}}{2}|\mathbf{b}|^2$
Thus, the magnitude of the numerator is:
$\left|\overrightarrow{CA} \times \mathbf{d}\right| = \frac{2}{3} \left( \frac{3\sqrt{3}}{2}|\mathbf{b}|^2 \right) = \sqrt{3}|\mathbf{b}|^2$

6. Calculate the final shortest distance.

Substituting the magnitude of the cross product and the magnitude of the direction vector into the distance formula:
$d = \frac{\sqrt{3}|\mathbf{b}|^2}{\sqrt{7}|\mathbf{b}|} = \sqrt{\frac{3}{7}}|\mathbf{b}| = \frac{\sqrt{21}}{7}|\mathbf{b}|$