平面 $p$ 和 $q$ 互相垂直。平面 $p$ 的方程为 $2x - z = 10$。平面 $q$ 包含具有如下方程的直线 $l$： $\mathbf{r} = \begin{pmatrix}5 \\ -3 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix}2 \\ -2 \\ 1 \end{pmatrix},$ 其中 $\lambda$ 是一个参数。

1. Find the normal vector to plane $q$ by taking the cross product of the normal vector of $p$ and the direction vector of $l$.

Plane $p$: $2x - z = 10$. Its normal vector is $\mathbf{n}_p = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}$.
Line $l$ is on plane $q$, so its direction vector $\mathbf{d}_l = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$ is parallel to $q$.
Since plane $q$ is perpendicular to plane $p$, $\mathbf{n}_p$ is also parallel to $q$.
The normal vector to $q$, $\mathbf{n}_q$, is perpendicular to both $\mathbf{n}_p$ and $\mathbf{d}_l$:
$\mathbf{n}_q = \mathbf{n}_p \times \mathbf{d}_l = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} \times \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 0(1) - (-1)(-2) \\ (-1)(2) - (2)(1) \\ 2(-2) - 0(2) \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ -4 \end{pmatrix}$
Taking a simpler parallel normal vector, let $\mathbf{n} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$.

2. Determine the Cartesian equation of plane $q$ using a point on line $l$.

Since line $l$ lies on plane $q$, the point $\mathbf{a} = \begin{pmatrix} 5 \\ -3 \\ 4 \end{pmatrix}$ lies on $q$.
The equation of $q$ is $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$:
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -3 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = 5(1) + (-3)(2) + 4(2)$
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = 5 - 6 + 8 = 7$
Converting to Cartesian form:
$x + 2y + 2z = 7 \text{ (shown)}$

1. Find the direction vector of the line of intersection by taking the cross product of the planes' normal vectors.

Let the normal vectors of planes $p$ and $q$ be $\mathbf{n}_p$ and $\mathbf{n}_q$ respectively.
$\mathbf{n}_p = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}, \quad \mathbf{n}_q = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$

The direction vector of the line of intersection $m$, denoted as $\mathbf{d}_m$, is perpendicular to both normal vectors:
$\mathbf{d}_m = \mathbf{n}_p \times \mathbf{n}_q = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} (0)(2) - (-1)(2) \\ (-1)(1) - (2)(2) \\ (2)(2) - (0)(1) \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \\ 4 \end{pmatrix}$

2. Find a common point that lies on both planes to serve as the position vector of the line.

To find a specific point on both planes, substitute an arbitrary value for one coordinate and solve the resulting simultaneous equations. Let $z = 0$:

From the equation of plane $p$ ($2x - z = 10$):
$2x - 0 = 10 \implies x = 5$

Substitute $x = 5$ and $z = 0$ into the equation of plane $q$ ($x + 2y + 2z = 7$):
$5 + 2y + 2(0) = 7 \implies 2y = 2 \implies y = 1$

Thus, a point on line $m$ is $(5, 1, 0)$.

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

Using the position vector of the point $(5, 1, 0)$ and the direction vector $\begin{pmatrix} 2 \\ -5 \\ 4 \end{pmatrix}$, we can write the equation of line $m$:
$\mathbf{r} = \begin{pmatrix} 5 \\ 1 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -5 \\ 4 \end{pmatrix}, \quad \mu \in \mathbb{R}$

1. Form the vector equation of the line passing through $A$ and perpendicular to plane $q$.

From part (a), the normal vector to plane $q$ is $\mathbf{n}_q = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$.
The line passing through $A(2, 1, -6)$ and perpendicular to $q$ has equation:
$l_{AF}: \mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ -6 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}, \text{ where } \mu \in \mathbb{R}$

2. Substitute the line equation into the cartesian equation of plane $q$ to find the parameter for $F$.

Since the foot of the perpendicular $F$ lies on $q$, its coordinates $(x, y, z)$ must satisfy $x + 2y + 2z = 7$:
$(2 + \mu) + 2(1 + 2\mu) + 2(-6 + 2\mu) = 7$
$2 + \mu + 2 + 4\mu - 12 + 4\mu = 7$
$9\mu - 8 = 7$
$9\mu = 15 \implies \mu = \frac{5}{3}$

3. Find the position vector of the foot of perpendicular $F$.

Substitute $\mu = \frac{5}{3}$ into the line equation:
$\vec{OF} = \begin{pmatrix} 2 \\ 1 \\ -6 \end{pmatrix} + \frac{5}{3} \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$
$\vec{OF} = \begin{pmatrix} 2 + 5/3 \\ 1 + 10/3 \\ -6 + 10/3 \end{pmatrix} = \begin{pmatrix} 11/3 \\ 13/3 \\ -8/3 \end{pmatrix}$

4. Calculate the distance $k$ using the vector $\vec{AF}$.

The vector $\vec{AF} = \frac{5}{3} \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$.
The shortest distance $k$ is the magnitude of $\vec{AF}$:
$k = |\vec{AF}| = \frac{5}{3} \sqrt{1^2 + 2^2 + 2^2}$
$k = \frac{5}{3}(3) = 5$

1. Determine the direction vector of the required lines.

Since the lines lie in plane $p$ and have a constant distance to plane $q$, they must be parallel to the intersection of $p$ and $q$.
The direction of this intersection is given by the cross product of their normals:
$\mathbf{d} = \mathbf{n}_p \times \mathbf{n}_q = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \\ 4 \end{pmatrix}$

2. State the equation of the first line.

From earlier parts, point $A(2, 1, -6)$ lies in $p$ and has a shortest distance of $k=5$ to $q$.
Thus, one such line must pass through $A$.
Equation of the first line:
$l_1: \mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ -6 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -5 \\ 4 \end{pmatrix}, \quad \lambda \in \mathbb{R}$

3. Use the distance formula to find the condition for any point on the lines.

For any point $(x,y,z)$ on these lines, the shortest distance to plane $q$ ($x+2y+2z=7$) must be $5$.
Using the perpendicular distance formula:
$\frac{|x + 2y + 2z - 7|}{\sqrt{1^2 + 2^2 + 2^2}} = 5$
$\frac{|x + 2y + 2z - 7|}{3} = 5$
$|x + 2y + 2z - 7| = 15$
$x + 2y + 2z = 22 \quad \text{or} \quad x + 2y + 2z = -8$
(Note: Point $A$ satisfies $x + 2y + 2z = -8$, confirming it lies on $l_1$)

4. Find a point on the second line and write its equation.

The second line is the intersection of plane $p$ ($2x - z = 10$) and the plane $x + 2y + 2z = 22$.
We find a specific point on this line by letting $x = 4$:
$z = 2(4) - 10 = -2$
Substitute $x=4$ and $z=-2$ into the second plane:
$4 + 2y + 2(-2) = 22 \implies 2y = 22 \implies y = 11$
A point on the second line is $(4, 11, -2)$.
Equation of the second line:
$l_2: \mathbf{r} = \begin{pmatrix} 4 \\ 11 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -5 \\ 4 \end{pmatrix}, \quad \mu \in \mathbb{R}$