在图中，点 $O$ 是直角棱锥顶角为 $V$、底面为长方形 $ABCD$ 的中心。相互垂直的单位向量 i, j, k 分别平行于 AB, BC, OV。$AB$、$BC$ 和 $OV$ 的长度分别为 $12$ 单位、$6$ 单位和 $6$ 单位。点 $M$ 是 $CV$ 的中点，点 $O$ 作为位置向量的原点。

1. Determine the position vectors of vertices A, C, and V.

The centre of the rectangle is at the origin $O$.
With dimensions $AB = 12$ and $BC = 6$, and taking note of the unit vectors directions, the position vectors of the corners are:
$\mathbf{OA} = \begin{pmatrix} -6 \\ -3 \\ 0 \end{pmatrix}, \quad \mathbf{OC} = \begin{pmatrix} 6 \\ 3 \\ 0 \end{pmatrix}$
Since $OV = 6$ and is parallel to $\mathbf{k}$:
$\mathbf{OV} = \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix}$

2. Find the position vector of point M.

Since $M$ is the mid-point of $CV$:
$\mathbf{OM} = \frac{\mathbf{OC} + \mathbf{OV}}{2} = \frac{1}{2} \left[ \begin{pmatrix} 6 \\ 3 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix} \right] = \begin{pmatrix} 3 \\ 1.5 \\ 3 \end{pmatrix}$

3. Find the direction vector of the line AM.

The direction of the line $AM$ is given by the displacement vector $\mathbf{AM}$:
$\mathbf{AM} = \mathbf{OM} - \mathbf{OA} = \begin{pmatrix} 3 \\ 1.5 \\ 3 \end{pmatrix} - \begin{pmatrix} -6 \\ -3 \\ 0 \end{pmatrix} = \begin{pmatrix} 9 \\ 4.5 \\ 3 \end{pmatrix}$

4. Formulate the equation of the line in the required form.

To match the required form with integer components, observe that $\mathbf{AM}$ is a scalar multiple of the given direction vector:
$\mathbf{AM} = 1.5 \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}$
Thus, a valid direction vector for the line is $\begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}$.
Using the position vector of point $A(-6, -3, 0)$, the vector equation of the line $AM$ is:
$\mathbf{r} = \begin{pmatrix} -6 \\ -3 \\ 0 \end{pmatrix} + t \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix} \quad \text{(shown)}$

1. Find the displacement vector $\vec{AB}$ and identify the direction vector of the line AM.

From the coordinates derived in the context, $\mathbf{a} = \begin{pmatrix} -6 \\ -3 \\ 0 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 6 \\ -3 \\ 0 \end{pmatrix}$.

$\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 6 \\ -3 \\ 0 \end{pmatrix} - \begin{pmatrix} -6 \\ -3 \\ 0 \end{pmatrix} = \begin{pmatrix} 12 \\ 0 \\ 0 \end{pmatrix}$.

From part (i), the direction vector of line AM is $\mathbf{d} = \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}$.

2. Calculate the cross product of $\vec{AB}$ and the direction vector $\mathbf{d}$.

$\vec{AB} \times \mathbf{d} = \begin{pmatrix} 12 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix} = \begin{pmatrix} (0)(2) - (0)(3) \\ (0)(6) - (12)(2) \\ (12)(3) - (0)(6) \end{pmatrix} = \begin{pmatrix} 0 \\ -24 \\ 36 \end{pmatrix}$

3. Compute the perpendicular distance using the magnitude of the cross product.

The perpendicular distance from $B$ to line AM is given by $D = \frac{|\vec{AB} \times \mathbf{d}|}{|\mathbf{d}|}$.

$|\vec{AB} \times \mathbf{d}| = \sqrt{0^2 + (-24)^2 + 36^2} = \sqrt{1872} = 12\sqrt{13}$
$|\mathbf{d}| = \sqrt{6^2 + 3^2 + 2^2} = \sqrt{49} = 7$

$D = \frac{12\sqrt{13}}{7} \approx 6.18 \text{ units}$

1. Find a normal vector to the plane AMB.

The direction vector of line $AM$ is $\mathbf{d}_1 = \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}$ (from part (i)).
Since $A = (-6, -3, 0)$ and $B = (6, -3, 0)$, the vector $\vec{AB} = \begin{pmatrix} 12 \\ 0 \\ 0 \end{pmatrix}$. A parallel, simplified direction vector is $\mathbf{d}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$.
The normal vector to plane $AMB$, $\mathbf{n}$, is the cross product of these two vectors:
$\mathbf{n} = \mathbf{d}_2 \times \mathbf{d}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -2 \\ 3 \end{pmatrix}$

2. Find the direction vector of the line DV.

Given $D = (-6, 3, 0)$ and $V = (0, 0, 6)$, the direction vector of line $DV$ is:
$\vec{DV} = \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix} - \begin{pmatrix} -6 \\ 3 \\ 0 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \\ 6 \end{pmatrix}$
We can factor out 3 to use a simpler parallel direction vector $\mathbf{m} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$.

3. Calculate the acute angle between the line DV and the plane AMB.

Let the acute angle between the line $DV$ and the plane $AMB$ be $\theta$. Using the relationship between a line and a plane's normal:
$\sin \theta = \frac{|\mathbf{m} \cdot \mathbf{n}|}{|\mathbf{m}| |\mathbf{n}|}$
$\mathbf{m} \cdot \mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ -2 \\ 3 \end{pmatrix} = 0 + 2 + 6 = 8$
$|\mathbf{m}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$
$|\mathbf{n}| = \sqrt{0^2 + (-2)^2 + 3^2} = \sqrt{13}$
$\sin \theta = \frac{8}{3\sqrt{13}}$
$\theta = \sin^{-1}\left( \frac{8}{3\sqrt{13}} \right) \approx 47.697^\circ$
The acute angle is $47.7^\circ$ (to 1 d.p.).

1. Relate the intersection of the three planes to the line $AM$ and the plane $\Pi$.

The planes $AMB$ and $AMD$ intersect at the line $AM$. For the three planes $AMB$, $AMD$ and $\Pi$ to have no point in common, the line of intersection $AM$ must be parallel to the plane $\Pi$ (and not lie on it).

2. Use the orthogonality of the line's direction vector and the plane's normal vector to solve for $a$.

From part (i), the direction vector of line $AM$ is $\mathbf{d} = \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}$, and the normal vector of plane $\Pi$ is $\mathbf{n} = \begin{pmatrix} -1 \\ 4 \\ a \end{pmatrix}$.

For the line to be parallel to the plane, their dot product must be zero ($\mathbf{d} \cdot \mathbf{n} = 0$):
$\begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 4 \\ a \end{pmatrix} = 0$
$-6 + 12 + 2a = 0$
$6 + 2a = 0$
$2a = -6$
$a = -3$

(Note: Checking point $A(-6, -3, 0)$ in the plane equation gives $6 - 12 + 0 = -6 \neq 4$, confirming the line $AM$ does not lie on $\Pi$.)