一家钻探公司计划在山中铺设一条直管道 $AB$。点 $(x, y, z)$ 是相对于山脚下坐标为 $(0, 0, 0)$ 的主控制中心定义的，其中单位为米。$x$ 轴指向东，$y$ 轴指向北，$z$ 轴垂直向上。点 $A$ 的坐标为 $(-200, 150, 10)$，点 $B$ 的坐标为 $(100, 10, a)$，其中 $a$ 是一个整数。点 $B$ 的海拔高于点 $A$。

1. Set up an equation using the 3D distance formula for $AB$

Using the coordinates $A(-200, 150, 10)$ and $B(100, 10, a)$, we apply the distance formula:
$|AB|^2 = (x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2$
$337^2 = (100 - (-200))^2 + (10 - 150)^2 + (a - 10)^2$
$113569 = 300^2 + (-140)^2 + (a - 10)^2$

2. Solve the resulting quadratic equation for $a$

$113569 = 90000 + 19600 + (a - 10)^2$
$113569 = 109600 + (a - 10)^2$
$(a - 10)^2 = 3969$
$a - 10 = \pm \sqrt{3969}$
$a - 10 = \pm 63$
$a = 73 \quad \text{or} \quad a = -53$

3. Determine the correct value of $a$ using the context and state the coordinates of $B$

We are given that Point $B$ is at a higher altitude than Point $A$. This means the $z$-coordinate of $B$ must be greater than the $z$-coordinate of $A$ ($a > 10$).
Therefore, we reject $a = -53$ and accept $a = 73$.

The coordinates of $B$ are $(100, 10, 73)$.

1. Find the coordinates of $B$ using the given length of the pipeline $AB$ (from part i).

Given $A(-200, 150, 10)$ and $B(100, 10, a)$:
$AB^2 = (100 - (-200))^2 + (10 - 150)^2 + (a - 10)^2$
$337^2 = 300^2 + (-140)^2 + (a - 10)^2$
$113569 = 90000 + 19600 + (a - 10)^2$
$(a - 10)^2 = 3969$
$a - 10 = \pm 63$
Since $B$ is at a higher altitude than $A$, $a > 10$. Thus, $a - 10 = 63 \implies a = 73$.
Coordinates of $B$ are $(100, 10, 73)$.

2. Form the vector equation of the line passing through $A$ and $B$.

$\vec{AB} = \begin{pmatrix} 100 \\ 10 \\ 73 \end{pmatrix} - \begin{pmatrix} -200 \\ 150 \\ 10 \end{pmatrix} = \begin{pmatrix} 300 \\ -140 \\ 63 \end{pmatrix}$
The vector equation of the line $AB$ is:
$\mathbf{r} = \begin{pmatrix} -200 \\ 150 \\ 10 \end{pmatrix} + \lambda \begin{pmatrix} 300 \\ -140 \\ 63 \end{pmatrix}, \lambda \in \mathbb{R}$

3. Substitute the parametric form of the line into the Cartesian equation of the plane to find the parameter $\lambda$.

The equation of the plane is $20x + y + 2z = -837$. Substituting the coordinates from the line equation:
$20(-200 + 300\lambda) + (150 - 140\lambda) + 2(10 + 63\lambda) = -837$
$-4000 + 6000\lambda + 150 - 140\lambda + 20 + 126\lambda = -837$
$(-4000 + 150 + 20) + (6000 - 140 + 126)\lambda = -837$
$-3830 + 5986\lambda = -837$
$5986\lambda = 2993$
$\lambda = \frac{1}{2}$

4. Calculate the final coordinates of the intersection point.

Substitute $\lambda = \frac{1}{2}$ back into the line equation:
$\mathbf{r} = \begin{pmatrix} -200 \\ 150 \\ 10 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 300 \\ -140 \\ 63 \end{pmatrix}$
$\mathbf{r} = \begin{pmatrix} -200 + 150 \\ 150 - 70 \\ 10 + 31.5 \end{pmatrix} = \begin{pmatrix} -50 \\ 80 \\ 41.5 \end{pmatrix}$
The coordinates are $(-50, 80, 41.5)$.

1. Find the unknown coordinate $a$ for point $B$ using the given length of $AB$, and determine the direction vector $\vec{AB}$.

Using the condition from part (i) that $AB = 337$:
$\sqrt{(100 - (-200))^2 + (10 - 150)^2 + (a - 10)^2} = 337$
$300^2 + (-140)^2 + (a - 10)^2 = 113569$
$109600 + (a - 10)^2 = 113569 \implies (a - 10)^2 = 3969$
Since $B$ is at a higher altitude than $A$ ($z=10$), $a > 10$. Thus, $a - 10 = 63 \implies a = 73$.
So, $B$ is $(100, 10, 73)$.
The vector $\vec{AB}$ is:
$\vec{AB} = \begin{pmatrix} 100 \\ 10 \\ 73 \end{pmatrix} - \begin{pmatrix} -200 \\ 150 \\ 10 \end{pmatrix} = \begin{pmatrix} 300 \\ -140 \\ 63 \end{pmatrix}$

2. Identify that minimum cable lengths correspond to perpendicular distances, meaning $PQ$ is the length of the orthogonal projection of $\vec{AB}$ onto the plane.

For the minimum length of cable to be used, the cables $AP$ and $BQ$ must be perpendicular to the layer of rock. Therefore, $P$ and $Q$ are the feet of the perpendiculars from $A$ and $B$, making $PQ$ the length of the orthogonal projection of $\vec{AB}$ onto the plane.

The normal vector to the plane $20x + y + 2z = -837$ is $\mathbf{n} = \begin{pmatrix} 20 \\ 1 \\ 2 \end{pmatrix}$.
The length of the projection of $\vec{AB}$ onto the normal $\mathbf{n}$ is given by:
$d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{\left| \begin{pmatrix} 300 \\ -140 \\ 63 \end{pmatrix} \cdot \begin{pmatrix} 20 \\ 1 \\ 2 \end{pmatrix} \right|}{\sqrt{20^2 + 1^2 + 2^2}}$
$d = \frac{|6000 - 140 + 126|}{\sqrt{405}} = \frac{5986}{\sqrt{405}}$

3. Use Pythagoras' theorem on the orthogonal vector components to find the length $PQ$.

The vector $\vec{AB}$ can be resolved into a component parallel to the plane (of length $PQ$) and a component perpendicular to the plane (of length $d$). By Pythagoras' theorem:
$PQ = \sqrt{|\vec{AB}|^2 - d^2}$
Given $|\vec{AB}| = 337$:
$PQ = \sqrt{337^2 - \left(\frac{5986}{\sqrt{405}}\right)^2}$
$PQ = \sqrt{113569 - \frac{35832196}{405}} = \sqrt{25094.44...} \approx 158.41$
Therefore, the required minimum length $PQ$ is $158 \text{ m}$ (to 3 s.f.).

1. Calculate the missing $z$-coordinate of point $B$ to find the direction vector of the pipeline.

Using the distance formula for $AB = 337$:
$\sqrt{(100 - (-200))^2 + (10 - 150)^2 + (a - 10)^2} = 337$
$300^2 + (-140)^2 + (a - 10)^2 = 337^2$
$109600 + (a - 10)^2 = 113569$
$(a - 10)^2 = 3969 \implies a - 10 = 63$ (since $B$ is higher than $A$, $a > 10$)
Thus, $a = 73$.

The direction vector of the pipeline is $\vec{AB} = \begin{pmatrix} 100 - (-200) \\ 10 - 150 \\ 73 - 10 \end{pmatrix} = \begin{pmatrix} 300 \\ -140 \\ 63 \end{pmatrix}$.

2. Find the angle between the pipeline and the horizontal plane using the normal to the plane.

The horizontal plane (the $xy$-plane) has a normal vector $\mathbf{n} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

Let the angle between the pipeline and the horizontal plane be $\theta$.
$\sin \theta = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\vec{AB}| |\mathbf{n}|}$
$\sin \theta = \frac{\left| \begin{pmatrix} 300 \\ -140 \\ 63 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right|}{337 \times 1}$
$\sin \theta = \frac{63}{337}$
$\theta = \sin^{-1}\left(\frac{63}{337}\right) \approx 10.772^\circ$
$\theta = 10.8^\circ$ (shown)

1. Evaluate the rate of altitude loss at $t = 10$.

Let $h$ be the altitude lost.
The rate of altitude loss is given by:
$\frac{\mathrm{d}h}{\mathrm{d}t} = 0.3t$
At $t = 10$ seconds,
$\frac{\mathrm{d}h}{\mathrm{d}t} = 0.3(10) = 3 \text{ m/s}$

2. Establish the relationship between the distance travelled along the pipeline and the altitude lost.

Let $s$ be the distance travelled along the pipeline from point $B$, and $\theta$ be the angle of the pipeline to the horizontal.
From the coordinates of $A(-200, 150, 10)$ and finding $B(100, 10, 73)$ in part (i), the total vertical change in altitude is $73 - 10 = 63\text{ m}$ over the full pipeline length of $337\text{ m}$.
Therefore, $\sin \theta = \frac{63}{337}$.
Using the right-angled triangle formed by the pipeline's slope:
$\sin \theta = \frac{h}{s} \implies s = \frac{h}{\sin \theta} = \frac{337}{63}h$

3. Apply the chain rule to find the speed along the pipeline.

Differentiating $s$ with respect to $h$ gives $\frac{\mathrm{d}s}{\mathrm{d}h} = \frac{337}{63}$.
Using the chain rule for connected rates of change:
$\frac{\mathrm{d}s}{\mathrm{d}t} = \frac{\mathrm{d}s}{\mathrm{d}h} \times \frac{\mathrm{d}h}{\mathrm{d}t}$
$\frac{\mathrm{d}s}{\mathrm{d}t} = \frac{337}{63} \times 3 = \frac{337}{21} \approx 16.048 \text{ m/s}$
The speed is $16.0 \text{ m/s}$ (to 3 s.f.).