平面 $p$ 通过点 $A$、$B$ 和 $C$，它们的坐标分别为 $(0, 1, 1)$、$(2, -1, 4)$ 和 $(-2, -1, 0)$。

1. Find two direction vectors parallel to the plane.

$\vec{AB} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix}$

$\vec{AC} = \begin{pmatrix} -2 \\ -1 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}$

2. Calculate the normal vector to the plane using the cross product.

$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} \times \begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} (-2)(-1) - (3)(-2) \\ (3)(-2) - (2)(-1) \\ (2)(-2) - (-2)(-2) \end{pmatrix} = \begin{pmatrix} 8 \\ -4 \\ -8 \end{pmatrix}$

A simplified normal vector is $\begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}$.

3. Determine the cartesian equation of the plane.

The equation of the plane is $\mathbf{r} \cdot \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix} = \mathbf{a} \cdot \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}$.

Substituting point $A(0, 1, 1)$:
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix} = 0(2) + 1(-1) + 1(-2) = -3$

Thus, the cartesian equation is $2x - y - 2z = -3$ (shown).

1. Identify the direction vector of the line $l$ and the normal vector of the plane $p$.

From the given equation of line $l$, the direction vector is $\mathbf{d} = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$.

From the cartesian equation of plane $p$ derived in part (i), $2x - y - 2z = -3$, the normal vector is $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}$.

2. Use the scalar product to find the sine of the acute angle between the line and the plane.

Let $\theta$ be the acute angle between the line $l$ and the plane $p$.
The relationship is given by $\sin \theta = \dfrac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}$.

$\sin \theta = \frac{\left| \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix} \right|}{\sqrt{1^2 + 0^2 + (-2)^2} \sqrt{2^2 + (-1)^2 + (-2)^2}}$

$\sin \theta = \frac{|1(2) + 0(-1) + (-2)(-2)|}{\sqrt{5} \sqrt{9}}$

$\sin \theta = \frac{6}{3\sqrt{5}} = \frac{2}{\sqrt{5}}$

3. Calculate the acute angle.

$\theta = \sin^{-1}\left( \frac{2}{\sqrt{5}} \right) \approx 63.435^\circ$

Rounding to 1 decimal place, the acute angle is $63.4^\circ$.

1. Equate the position vector of $Q$ to the equation of the line $l$.

$\begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$

2. Solve for $\lambda$ using the components and verify consistency.

From the $x$-component:
$5 = 2 + \lambda \implies \lambda = 3$

Check with the $y$-component:
$-1 = -1 + \lambda(0) \implies -1 = -1$ (consistent)

Check with the $z$-component:
$-2 = 4 - 2\lambda \implies 2\lambda = 6 \implies \lambda = 3$ (consistent)

Since $\lambda = 3$ satisfies all components, $Q$ lies on $l$.

1. Formulate the equation of the line passing through $Q$ and perpendicular to the plane $p$.

The normal vector to the plane $p$ is $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}$.
Let the line passing through $Q$ and perpendicular to $p$ be $l_Q$.
The vector equation of $l_Q$ is:
$\mathbf{r} = \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}, \mu \in \mathbb{R}$

2. Find the value of the parameter $\mu$ at the point of intersection with the plane $p$.

Substitute the parametric coordinates $x = 5 + 2\mu$, $y = -1 - \mu$, and $z = -2 - 2\mu$ into the cartesian equation of the plane $2x - y - 2z = -3$:
$2(5 + 2\mu) - (-1 - \mu) - 2(-2 - 2\mu) = -3$
$10 + 4\mu + 1 + \mu + 4 + 4\mu = -3$
$15 + 9\mu = -3$
$9\mu = -18 \implies \mu = -2$

3. Calculate the coordinates of the foot of the perpendicular, $F$.

Substitute $\mu = -2$ into the equation of line $l_Q$:
$\vec{OF} = \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} - 2 \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$
The foot of the perpendicular is $F(1, 1, 2)$.

4. Calculate the perpendicular distance from $Q$ to $F$.

$QF = \sqrt{(5 - 1)^2 + (-1 - 1)^2 + (-2 - 2)^2}$
$QF = \sqrt{4^2 + (-2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$

5. Use Pythagoras' theorem to find the distance $FR$.

Since $F$ is the foot of the perpendicular from $Q$ to the plane $p$, $\vec{QF}$ is perpendicular to any vector on the plane, including $\vec{FR}$.
Thus, $\triangle QFR$ is a right-angled triangle at $F$.
$FR^2 + QF^2 = QR^2$
Given $QR = \sqrt{45}$:
$FR^2 + 6^2 = (\sqrt{45})^2$
$FR^2 + 36 = 45$
$FR^2 = 9 \implies FR = 3$

6. Describe geometrically the locus of $R$.

The point $R$ lies on the plane $p$ and remains at a constant distance of $3$ from a fixed point $F(1, 1, 2)$ which also lies on the plane.
Therefore, the locus of $R$ is a circle on the plane $p$ with centre $(1, 1, 2)$ and radius $3$.

1. Identify the point of intersection between line $l$ and plane $p$.

Line $l$ has the equation $\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$.
The position vector $\begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}$ corresponds to point $B$, which is given in the question to lie on the plane $p$.
Hence, $B(2, -1, 4)$ is the point of intersection between line $l$ and plane $p$.

2. Find the foot of perpendicular $F$ from $Q$ to plane $p$ (deriving result from part iv).

The line passing through $Q(5, -1, -2)$ and perpendicular to plane $p$ has direction vector $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}$.
Its vector equation is $\mathbf{r} = \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}$.
Substituting into the cartesian equation of $p$ ($2x - y - 2z = -3$):
$2(5 + 2\mu) - (-1 - \mu) - 2(-2 - 2\mu) = -3$
$10 + 4\mu + 1 + \mu + 4 + 4\mu = -3$
$9\mu + 15 = -3 \implies \mu = -2$
Thus, the foot of the perpendicular $F$ is:
$\vec{OF} = \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} - 2\begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$

3. Find the position vector of $Q'$, the reflection of $Q$ in $p$.

Let $Q'$ be the reflection of $Q$ in $p$. The foot of the perpendicular $F$ is the midpoint of $QQ'$.
$\vec{OF} = \frac{\vec{OQ} + \vec{OQ'}}{2}$
$\vec{OQ'} = 2\vec{OF} - \vec{OQ} = 2\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} -3 \\ 3 \\ 6 \end{pmatrix}$

4. Form the vector equation of the reflected line.

The reflected line passes through the point of intersection $B$ and the reflected point $Q'$.
The direction vector of this reflected line is:
$\vec{BQ'} = \vec{OQ'} - \vec{OB} = \begin{pmatrix} -3 \\ 3 \\ 6 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} -5 \\ 4 \\ 2 \end{pmatrix}$
The vector equation of the reflected line is:
$\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + \gamma \begin{pmatrix} -5 \\ 4 \\ 2 \end{pmatrix}, \quad \gamma \in \mathbb{R}$