A player is controlling an avatar in a virtual reality game where points $(x,y,z)$ are defined relative to a fixed point $O$. Units are measured in metres and we assume that the avatar moves in a straight line from point to point. Initially the avatar is at a point $A$ with coordinates $(-3,4,-1)$. The avatar first travels in the direction parallel to $\mathbf{i}-3\mathbf{j}$ towards a surface $p_1$, with equation $x+y+3=0$. The avatar is supposed to destroy the target at a point $B$ with coordinates $(-1,-2,-1)$ so that it could pass through the surface $p_1$ and move towards its final destination at a point $C$.

1. Substitute the coordinates of $B$ into the equation of $p_1$.

The coordinates of $B$ are $(-1, -2, -1)$. Therefore, $x = -1$ and $y = -2$.
Substitute these into the left-hand side of the equation of $p_1$ ($x + y + 3 = 0$):
$LHS = -1 - 2 + 3$
$LHS = 0 = RHS$
Since the equation is satisfied, $B$ lies on $p_1$.

1. Identify the normal vector to the plane and formulate a displacement vector from a point on the plane to $A$.

From the equation of $p_1$, $x+y+3=0$, the normal vector to the plane is $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$.

From part (i), we know the target point $B(-1, -2, -1)$ lies on $p_1$. The displacement vector from $B$ to $A$ is:
$\vec{BA} = \vec{OA} - \vec{OB} = \begin{pmatrix} -3 \\ 4 \\ -1 \end{pmatrix} - \begin{pmatrix} -1 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \\ 0 \end{pmatrix}$

2. Calculate the shortest distance by finding the length of projection of $\vec{BA}$ onto the normal vector.

The shortest distance $d$ from point $A$ to the plane $p_1$ is the length of projection of $\vec{BA}$ onto the normal vector $\mathbf{n}$:
$d = \frac{|\vec{BA} \cdot \mathbf{n}|}{|\mathbf{n}|}$
$d = \frac{\left| \begin{pmatrix} -2 \\ 6 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \right|}{\sqrt{1^2 + 1^2 + 0^2}}$
$d = \frac{|(-2)(1) + (6)(1) + (0)(0)|}{\sqrt{2}}$
$d = \frac{4}{\sqrt{2}} = 2\sqrt{2}$

1. Identify the direction vector of the path $AB$ and the normal vector of the plane $p_1$.

From the problem description, the direction vector of the path is $\mathbf{d} = \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}$.

The Cartesian equation of $p_1$ is $x + y + 3 = 0$, so its normal vector is $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$.

2. Use the scalar product to find the sine of the angle $\theta$ between the path and the plane.

Let $\theta$ be the acute angle between the path $AB$ and $p_1$.
$\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}|\,|\mathbf{n}|}$
$\sin \theta = \frac{\left| \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \right|}{\sqrt{1^2 + (-3)^2 + 0^2} \sqrt{1^2 + 1^2 + 0^2}}$
$\sin \theta = \frac{|(1)(1) + (-3)(1) + (0)(0)|}{\sqrt{10}\sqrt{2}}$
$\sin \theta = \frac{|-2|}{\sqrt{20}} = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}}$

3. Calculate the acute angle.

$\theta = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \approx 26.565^\circ$
$\theta = 26.6^\circ \text{ (to 1 d.p.)}$

1. Find the position vector of the foot of the perpendicular, $F$, from $A$ to the surface $p_1$.

The normal vector to $p_1$ ($x+y+3=0$) is $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$.
The line passing through $A$ and perpendicular to $p_1$ has equation:
$\mathbf{r} = \begin{pmatrix} -3 \\ 4 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \lambda \in \mathbb{R}$
At the intersection with $p_1$, the coordinates of the line satisfy $x+y+3=0$:
$(-3+\lambda) + (4+\lambda) + 3 = 0$
$2\lambda + 4 = 0 \implies \lambda = -2$
Substituting $\lambda = -2$ gives the position vector of $F$:
$\vec{OF} = \begin{pmatrix} -3 \\ 4 \\ -1 \end{pmatrix} - 2\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -5 \\ 2 \\ -1 \end{pmatrix}$

2. Find the position vector of $A'$, the reflection of $A$ about $p_1$.

Since $F$ is the midpoint of the line segment $AA'$, we have $\vec{OF} = \frac{\vec{OA} + \vec{OA'}}{2}$.
$\vec{OA'} = 2\vec{OF} - \vec{OA}$
$\vec{OA'} = 2\begin{pmatrix} -5 \\ 2 \\ -1 \end{pmatrix} - \begin{pmatrix} -3 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} -10 \\ 4 \\ -2 \end{pmatrix} - \begin{pmatrix} -3 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} -7 \\ 0 \\ -1 \end{pmatrix}$

3. Determine the direction vector for the path $A'B$ and state the vector equation of the straight path.

The direction vector from $A'$ to $B$ is:
$\vec{A'B} = \vec{OB} - \vec{OA'} = \begin{pmatrix} -1 \\ -2 \\ -1 \end{pmatrix} - \begin{pmatrix} -7 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \\ 0 \end{pmatrix}$
We can scale this vector to obtain a simpler parallel direction vector, such as $\begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}$.
Using the position vector of $B$, the vector equation of the path is:
$\mathbf{r} = \begin{pmatrix} -1 \\ -2 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}, \mu \in \mathbb{R}$

1. Find the value of $k$ by using the perpendicular distance between the parallel planes $p_1$ and $p_2$.

The equation of surface $p_1$ is $x+y=-3$ and $p_2$ is $x+y=k$.
The planes are parallel with a common normal vector $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$.
The distance from any point on $p_1$ (e.g., target $B(-1, -2, -1)$) to $p_2$ is given by the distance formula to a plane:
$\text{Distance} = \frac{|(-1) + (-2) - k|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|-3-k|}{\sqrt{2}}$
Equating this to the given distance $35\sqrt{2}$:
$\frac{|-3-k|}{\sqrt{2}} = 35\sqrt{2}$
$|-3-k| = 70$
Since $k$ is a positive constant ($k > 0$):
$k+3 = 70 \implies k = 67$

2. Formulate the vector equation of the path taken by the avatar from $B$ to $C$.

The avatar moves from $B(-1, -2, -1)$ towards $C$ in the direction parallel to $3\mathbf{i}+\mathbf{j}+2\mathbf{k}$.
The equation of the line representing this path is:
$\mathbf{r} = \begin{pmatrix} -1 \\ -2 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix}, \quad \lambda > 0$

3. Substitute the line's parametric equations into the equation of $p_2$ to find the coordinates of $C$.

Since $C$ lies on $p_2: x+y=67$, we substitute the $x$ and $y$ coordinates of the line into the plane equation:
$(-1 + 3\lambda) + (-2 + \lambda) = 67$
$-3 + 4\lambda = 67$
$4\lambda = 70 \implies \lambda = 17.5$
The position vector of $C$ is found by substituting $\lambda = 17.5$ back into the line equation:
$\vec{OC} = \begin{pmatrix} -1 \\ -2 \\ -1 \end{pmatrix} + 17.5 \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -1 + 52.5 \\ -2 + 17.5 \\ -1 + 35 \end{pmatrix} = \begin{pmatrix} 51.5 \\ 15.5 \\ 34 \end{pmatrix}$
Thus, the coordinates of $C$ are $\left(\frac{103}{2}, \frac{31}{2}, 34\right)$.