以原点 $O$ 为基准，点 $A,B,C,D$ 和 $E$ 的坐标分别为 $A(2,3,4),\; B(6,5,7),\; C(8,9,6),\; D(4,7,3)$ 且 $E(5,6,10)$。

1. Find two non-parallel direction vectors that lie on the plane, such as $\vec{AB}$ and $\vec{AE}$.

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 6 \\ 5 \\ 7 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix}$
$\vec{AE} = \vec{OE} - \vec{OA} = \begin{pmatrix} 5 \\ 6 \\ 10 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 6 \end{pmatrix}$

2. Compute the cross product of the two direction vectors to find a normal vector to the plane.

$\mathbf{n} = \vec{AB} \times \vec{AE} = \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix} \times \begin{pmatrix} 3 \\ 3 \\ 6 \end{pmatrix} = \begin{pmatrix} (2)(6) - (3)(3) \\ (3)(3) - (4)(6) \\ (4)(3) - (2)(3) \end{pmatrix} = \begin{pmatrix} 3 \\ -15 \\ 6 \end{pmatrix}$
We can take a simpler normal vector $\mathbf{n_1}$ parallel to $\mathbf{n}$ by dividing by 3:
$\mathbf{n_1} = \begin{pmatrix} 1 \\ -5 \\ 2 \end{pmatrix}$

3. Formulate the Cartesian equation of the plane using $\mathbf{r} \cdot \mathbf{n_1} = \vec{OA} \cdot \mathbf{n_1}$.

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -5 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -5 \\ 2 \end{pmatrix}$
$x - 5y + 2z = 2(1) + 3(-5) + 4(2)$
$x - 5y + 2z = 2 - 15 + 8$
$x - 5y + 2z = -5 \text{ (shown)}$

1. Find the position vector of the midpoint $M$ of $EC$.

$\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OC} + \overrightarrow{OE}) = \frac{1}{2} \left[ \begin{pmatrix} 8 \\ 9 \\ 6 \end{pmatrix} + \begin{pmatrix} 5 \\ 6 \\ 10 \end{pmatrix} \right] = \begin{pmatrix} 13/2 \\ 15/2 \\ 8 \end{pmatrix}$

2. Find a direction vector for the line $DM$.

$\overrightarrow{DM} = \overrightarrow{OM} - \overrightarrow{OD} = \begin{pmatrix} 13/2 \\ 15/2 \\ 8 \end{pmatrix} - \begin{pmatrix} 4 \\ 7 \\ 3 \end{pmatrix} = \begin{pmatrix} 5/2 \\ 1/2 \\ 5 \end{pmatrix}$
By scaling the vector by a factor of $2$, we obtain an integer direction vector $\mathbf{d} = \begin{pmatrix} 5 \\ 1 \\ 10 \end{pmatrix}$.

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

$\mathbf{r} = \overrightarrow{OD} + \lambda \mathbf{d} = \begin{pmatrix} 4 \\ 7 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ 1 \\ 10 \end{pmatrix}, \text{ where } \lambda \in \mathbb{R}$

1. Identify the coordinates of the point $M$, which is the midpoint of $EC$.

Given $E(5,6,10)$ and $C(8,9,6)$, the midpoint $M$ is:
$M = \left( \frac{5+8}{2}, \frac{6+9}{2}, \frac{10+6}{2} \right) = \left( \frac{13}{2}, \frac{15}{2}, 8 \right)$

2. Construct the vector equation of the line passing through $M$ and perpendicular to the plane.

The normal vector to the plane $x - 5y + 2z = -5$ is $\mathbf{n} = \begin{pmatrix} 1 \\ -5 \\ 2 \end{pmatrix}$.

The line perpendicular to the plane passing through $M$ has equation:
$\mathbf{r} = \begin{pmatrix} 13/2 \\ 15/2 \\ 8 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -5 \\ 2 \end{pmatrix}, \quad \lambda \in \mathbb{R}$

3. Substitute the general point of the line into the plane equation to find the value of $\lambda$ at the intersection.

A general point on the line is $\left( \frac{13}{2} + \lambda, \frac{15}{2} - 5\lambda, 8 + 2\lambda \right)$. Substituting this into the plane equation $x - 5y + 2z = -5$:
$\left( \frac{13}{2} + \lambda \right) - 5\left( \frac{15}{2} - 5\lambda \right) + 2(8 + 2\lambda) = -5$
$\frac{13}{2} + \lambda - \frac{75}{2} + 25\lambda + 16 + 4\lambda = -5$
$30\lambda - 31 + 16 = -5$
$30\lambda - 15 = -5$
$30\lambda = 10 \implies \lambda = \frac{1}{3}$

4. Calculate the exact coordinates of the foot of the perpendicular.

Substitute $\lambda = \frac{1}{3}$ back into the coordinates of the general point:
$x = \frac{13}{2} + \frac{1}{3} = \frac{39+2}{6} = \frac{41}{6}$
$y = \frac{15}{2} - 5\left(\frac{1}{3}\right) = \frac{45-10}{6} = \frac{35}{6}$
$z = 8 + 2\left(\frac{1}{3}\right) = \frac{24+4}{3} = \frac{26}{3}$

The exact coordinates of the foot of the perpendicular are $\left(\frac{41}{6}, \frac{35}{6}, \frac{26}{3}\right)$.

1. Determine the vector $\vec{MF}$ using the coordinates of $M$ and the foot of the perpendicular $F$ derived in previous parts.

From earlier parts, the midpoint $M$ of $E(5,6,10)$ and $C(8,9,6)$ is $\left(\frac{13}{2}, \frac{15}{2}, 8\right)$.

The foot of the perpendicular $F$ is found by intersecting the normal line $\mathbf{r} = \begin{pmatrix} 13/2 \\ 15/2 \\ 8 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -5 \\ 2 \end{pmatrix}$ with the surface $x - 5y + 2z = -5$.
Solving gives $\lambda = \frac{1}{3}$, so $F$ has coordinates $\left(\frac{41}{6}, \frac{35}{6}, \frac{26}{3}\right)$.

Thus, the displacement vector is:
$\vec{MF} = \vec{OF} - \vec{OM} = \begin{pmatrix} 41/6 \\ 35/6 \\ 52/6 \end{pmatrix} - \begin{pmatrix} 39/6 \\ 45/6 \\ 48/6 \end{pmatrix} = \begin{pmatrix} 1/3 \\ -5/3 \\ 2/3 \end{pmatrix}$

2. Calculate the exact shortest distance by finding the magnitude of $\vec{MF}$.

The shortest distance from $M$ to the surface is the length of $\vec{MF}$:
$|\vec{MF}| = \sqrt{ \left(\frac{1}{3}\right)^2 + \left(-\frac{5}{3}\right)^2 + \left(\frac{2}{3}\right)^2 }$
$= \sqrt{ \frac{1}{9} + \frac{25}{9} + \frac{4}{9} } = \sqrt{\frac{30}{9}}$
$= \frac{\sqrt{30}}{3}$

1. Find the coordinates of $M$, state the equation of the line $DM$, and verify $P$ lies on it.

The coordinates of $M$, the midpoint of $E(5,6,10)$ and $C(8,9,6)$, are:
$M = \left(\frac{5+8}{2}, \frac{6+9}{2}, \frac{10+6}{2}\right) = (6.5, 7.5, 8)$
The direction vector of $DM$ is:
$\vec{DM} = \begin{pmatrix} 6.5 \\ 7.5 \\ 8 \end{pmatrix} - \begin{pmatrix} 4 \\ 7 \\ 3 \end{pmatrix} = \begin{pmatrix} 2.5 \\ 0.5 \\ 5 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 5 \\ 1 \\ 10 \end{pmatrix}$
The line $DM$ has equation $\mathbf{r} = \begin{pmatrix} 4 \\ 7 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ 1 \\ 10 \end{pmatrix}$.
Substituting $P(9,8,13)$ into the line equation:
$\begin{pmatrix} 9 \\ 8 \\ 13 \end{pmatrix} = \begin{pmatrix} 4 \\ 7 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ 1 \\ 10 \end{pmatrix} \implies \begin{pmatrix} 5 \\ 1 \\ 10 \end{pmatrix} = \lambda \begin{pmatrix} 5 \\ 1 \\ 10 \end{pmatrix} \implies \lambda = 1$
Since there is a consistent solution for $\lambda$, $P$ lies on the line $DM$.

2. Verify that $P$ lies on the given plane to conclude it is the point of intersection.

Substitute the coordinates of $P(9,8,13)$ into the Cartesian equation of the plane $x - 5y + 2z = -5$:
$\text{LHS} = (9) - 5(8) + 2(13) = 9 - 40 + 26 = -5$
Since $\text{LHS} = \text{RHS}$, $P$ lies on the plane.
Hence, the line $DM$ intersects the surface at $P(9,8,13)$.

3. Determine the position vector of $M'$, the reflection of $M$ about the plane.

To reflect the line, we can reflect $M$ about the plane. First, find the foot of the perpendicular $F$ from $M$ to the plane (as required in part c).
The line normal to the plane through $M$ is $\mathbf{r} = \begin{pmatrix} 6.5 \\ 7.5 \\ 8 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -5 \\ 2 \end{pmatrix}$.
Substitute into the plane equation:
$(6.5 + \mu) - 5(7.5 - 5\mu) + 2(8 + 2\mu) = -5 \implies 30\mu - 15 = -5 \implies \mu = \frac{1}{3}$
The reflection $M'$ of $M$ about the plane corresponds to $\mu = 2 \times \frac{1}{3} = \frac{2}{3}$:
$\vec{OM'} = \begin{pmatrix} 13/2 \\ 15/2 \\ 8 \end{pmatrix} + \frac{2}{3} \begin{pmatrix} 1 \\ -5 \\ 2 \end{pmatrix} = \begin{pmatrix} 43/6 \\ 25/6 \\ 28/3 \end{pmatrix}$.

4. Find the direction vector of the reflected line and state its vector equation.

The reflected line passes through the intersection point $P$ and the reflected point $M'$. Its direction vector is:
$\vec{M'P} = \vec{OP} - \vec{OM'} = \begin{pmatrix} 9 \\ 8 \\ 13 \end{pmatrix} - \begin{pmatrix} 43/6 \\ 25/6 \\ 56/6 \end{pmatrix} = \begin{pmatrix} 11/6 \\ 23/6 \\ 22/6 \end{pmatrix} = \frac{1}{6}\begin{pmatrix} 11 \\ 23 \\ 22 \end{pmatrix}$
Using $\begin{pmatrix} 11 \\ 23 \\ 22 \end{pmatrix}$ as the direction vector, the vector equation of the reflected line is:
$\mathbf{r} = \begin{pmatrix} 9 \\ 8 \\ 13 \end{pmatrix} + t \begin{pmatrix} 11 \\ 23 \\ 22 \end{pmatrix}, \text{ where } t \in \mathbb{R}$.