A frigate is stationed at position $F(1, 2, 0)$. Two submarines $S_1$ and $S_2$ are under the sea surface. Submarine $S_1$ is at position $A(-2, -1, -1)$ and travelling in a path parallel to vector $-3\mathbf{i} + 2\mathbf{j} - \mathbf{k}$. An enemy submarine $S_2$ is detected at position $B(3, 2, -2)$ travelling in a path parallel to vector $-2\mathbf{i} - 3\mathbf{j} + \mathbf{k}$.

1. Write down the vector equations for the paths of $S_1$ and $S_2$.

Let the paths be $l_1$ and $l_2$.
$l_1: \mathbf{r}_1 = \begin{pmatrix} -2 \\ -1 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} -3 \\ 2 \\ -1 \end{pmatrix}$, $\lambda \in \mathbb{R}$
$l_2: \mathbf{r}_2 = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} -2 \\ -3 \\ 1 \end{pmatrix}$, $\mu \in \mathbb{R}$

2. Set up a system of equations by assuming the lines intersect.

Equating $\mathbf{r}_1$ and $\mathbf{r}_2$, we get:
$-2 - 3\lambda = 3 - 2\mu \implies 3\lambda - 2\mu = -5$ --- (1)
$-1 + 2\lambda = 2 - 3\mu \implies 2\lambda + 3\mu = 3$ --- (2)
$-1 - \lambda = -2 + \mu \implies \lambda + \mu = 1$ --- (3)

3. Solve for the parameters and check for consistency.

From (3), $\mu = 1 - \lambda$.
Substitute into (2):
$2\lambda + 3(1 - \lambda) = 3$
$2\lambda + 3 - 3\lambda = 3 \implies \lambda = 0$
Thus, $\mu = 1 - 0 = 1$.
Substitute $\lambda = 0$ and $\mu = 1$ into the LHS of (1):
$3(0) - 2(1) = -2$
Since the LHS ($-2$) $\neq$ RHS ($-5$), the system of equations is inconsistent.
Therefore, the paths of the submarines do not intersect.

1. Formulate the vector equation for the path of $S_2$ and express the vector $\vec{FP}$ to a general point on this line.

Let $l_2$ be the line representing the path of $S_2$. The equation of $l_2$ is $\mathbf{r} = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} -2 \\ -3 \\ 1 \end{pmatrix}$, where $\mu \in \mathbb{R}$.

The position vector of a general point $P$ on $l_2$ is $\vec{OP} = \begin{pmatrix} 3 - 2\mu \\ 2 - 3\mu \\ -2 + \mu \end{pmatrix}$.

Given the position of the frigate $\vec{OF} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$, the vector $\vec{FP}$ is:
$\vec{FP} = \vec{OP} - \vec{OF} = \begin{pmatrix} 3 - 2\mu \\ 2 - 3\mu \\ -2 + \mu \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 - 2\mu \\ -3\mu \\ -2 + \mu \end{pmatrix}$

2. Apply the shortest distance condition using the dot product, since $\vec{FP}$ must be perpendicular to the path of $S_2$.

For $P$ to be the closest point to $F$, the vector $\vec{FP}$ must be perpendicular to the direction vector of $l_2$, which is $\mathbf{d}_2 = \begin{pmatrix} -2 \\ -3 \\ 1 \end{pmatrix}$.

Therefore, their scalar (dot) product must be zero: $\vec{FP} \cdot \mathbf{d}_2 = 0$.
$\begin{pmatrix} 2 - 2\mu \\ -3\mu \\ -2 + \mu \end{pmatrix} \cdot \begin{pmatrix} -2 \\ -3 \\ 1 \end{pmatrix} = 0$

3. Evaluate the scalar product and solve for the parameter $\mu$.

$-2(2 - 2\mu) - 3(-3\mu) + 1(-2 + \mu) = 0$
$-4 + 4\mu + 9\mu - 2 + \mu = 0$
$14\mu - 6 = 0$
$14\mu = 6 \implies \mu = \frac{6}{14} = \frac{3}{7}$

4. Substitute $\mu$ back into $\vec{OP}$ to find the coordinates of $P$.

$\vec{OP} = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} + \frac{3}{7} \begin{pmatrix} -2 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 - \frac{6}{7} \\ 2 - \frac{9}{7} \\ -2 + \frac{3}{7} \end{pmatrix} = \begin{pmatrix} 15/7 \\ 5/7 \\ -11/7 \end{pmatrix}$

The coordinates of point $P$ are $\left(\frac{15}{7}, \frac{5}{7}, -\frac{11}{7}\right)$.

1. Find two non-parallel vectors that lie on or are parallel to the plane $\pi$.

The plane $\pi$ contains the point $F(1, 2, 0)$ and the path of $S_2$, which passes through $B(3, 2, -2)$ with direction $\mathbf{d_2} = \begin{pmatrix} -2 \\ -3 \\ 1 \end{pmatrix}$.

A vector on the plane connecting $F$ and $B$ is:
$\vec{FB} = \vec{OB} - \vec{OF} = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix}$

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

The normal vector $\mathbf{n}$ is orthogonal to both $\vec{FB}$ and $\mathbf{d_2}$:
$\mathbf{n} = \vec{FB} \times \mathbf{d_2} = \begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix} \times \begin{pmatrix} -2 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 0(1) - (-2)(-3) \\ (-2)(-2) - 2(1) \\ 2(-3) - 0(-2) \end{pmatrix} = \begin{pmatrix} -6 \\ 2 \\ -6 \end{pmatrix}$

A simpler parallel normal vector is $\mathbf{n_1} = \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}$.

3. Determine the cartesian equation of the plane.

Using the point $F(1, 2, 0)$ and the scalar product equation $\mathbf{r} \cdot \mathbf{n_1} = \vec{OF} \cdot \mathbf{n_1}$:
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} = 3(1) + (-1)(2) + 3(0) = 1$

The cartesian equation is:
$3x - y + 3z = 1$

4. Calculate the acute angle between the plane $\pi$ and the $x$-$y$ plane.

The normal vector to the $x$-$y$ plane is $\mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.
Let $\theta$ be the acute angle between the two planes.
$\cos \theta = \frac{|\mathbf{n_1} \cdot \mathbf{k}|}{|\mathbf{n_1}| |\mathbf{k}|}$
$\cos \theta = \frac{\left| \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right|}{\sqrt{3^2 + (-1)^2 + 3^2} \sqrt{1^2}}$
$\cos \theta = \frac{3}{\sqrt{19}}$
$\theta = \cos^{-1}\left(\frac{3}{\sqrt{19}}\right) \approx 46.5^\circ \text{ (to 1 d.p.)}$

1. Retrieve or calculate the position vector of point $P$ from part (ii).

Let the path of $S_2$ be $\mathbf{r} = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} -2 \\ -3 \\ 1 \end{pmatrix}$.
The vector $\vec{FP}$ from $F(1, 2, 0)$ is $\begin{pmatrix} 2 - 2\mu \\ -3\mu \\ -2 + \mu \end{pmatrix}$.
Since $P$ is the closest point to $F$, $\vec{FP} \cdot \begin{pmatrix} -2 \\ -3 \\ 1 \end{pmatrix} = 0$.
$-2(2 - 2\mu) - 3(-3\mu) + 1(-2 + \mu) = 0 \implies 14\mu - 6 = 0 \implies \mu = \frac{3}{7}$.
Substituting $\mu = \frac{3}{7}$ gives the position vector $\vec{OP} = \begin{pmatrix} 15/7 \\ 5/7 \\ -11/7 \end{pmatrix}$.

2. Find the value of $k$ that minimizes the distance between $Q$ and $P$.

The position of $Q$ is given by $\vec{OQ} = \begin{pmatrix} 1 \\ 2 \\ -k \end{pmatrix}$.
$\vec{QP} = \vec{OP} - \vec{OQ} = \begin{pmatrix} 15/7 - 1 \\ 5/7 - 2 \\ -11/7 - (-k) \end{pmatrix} = \begin{pmatrix} 8/7 \\ -9/7 \\ k - 11/7 \end{pmatrix}$.
The square of the distance is $QP^2 = \left(\frac{8}{7}\right)^2 + \left(-\frac{9}{7}\right)^2 + \left(k - \frac{11}{7}\right)^2 = \frac{145}{49} + \left(k - \frac{11}{7}\right)^2$.
For $QP$ to be at a minimum, the squared term must be zero:
$k - \frac{11}{7} = 0 \implies k = \frac{11}{7}$.

3. Calculate the minimum distance and determine if $S_2$ will be damaged.

Substitute $k = \frac{11}{7}$ into the distance formula:
Minimum distance $QP = \sqrt{\frac{145}{49}} = \frac{\sqrt{145}}{7} \approx 1.72$ units.
Since the minimum distance of $1.72$ units is greater than the damage radius of $0.1$ units, $S_2$ will not be damaged by the depth charge.