The diagram above shows a cuboid $OABCDEFG$ with horizontal base $OABC$ where $OA = 6\text{ cm}$ and $AB = 4\text{ cm}$. $OD$, $AE$, $BF$ and $CG$ are vertical with height $10\text{ cm}$. The point $O$ is taken as the origin and vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ are unit vectors in the directions $OA$, $OC$ and $OD$ respectively. The point $P$ on $OA$ and the point $Q$ on $BF$ are such that $OP : PA = 2 : 1$ and $BQ : QF = 3 : 2$. Point $R$ is the midpoint of $AB$.

1. Find the position vectors of points $P$, $Q$, and $R$.

From the given dimensions and ratios, we determine the coordinates of the vertices:
$A = (6, 0, 0)$, $B = (6, 4, 0)$, $F = (6, 4, 10)$.
Since $OP : PA = 2 : 1$, $P$ is $\frac{2}{3}$ along $OA$:
$\vec{OP} = \frac{2}{3}\begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix}$
Since $BQ : QF = 3 : 2$, $Q$ is $\frac{3}{5}$ along $BF$:
$\vec{OQ} = \vec{OB} + \frac{3}{5}\vec{BF} = \begin{pmatrix} 6 \\ 4 \\ 0 \end{pmatrix} + \frac{3}{5}\begin{pmatrix} 0 \\ 0 \\ 10 \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \\ 6 \end{pmatrix}$
Since $R$ is the midpoint of $AB$:
$\vec{OR} = \frac{1}{2}(\vec{OA} + \vec{OB}) = \frac{1}{2} \left( \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 6 \\ 4 \\ 0 \end{pmatrix} \right) = \begin{pmatrix} 6 \\ 2 \\ 0 \end{pmatrix}$

2. Find two directional vectors on the plane $PQR$.

$\vec{PR} = \vec{OR} - \vec{OP} = \begin{pmatrix} 6 \\ 2 \\ 0 \end{pmatrix} - \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}$
$\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 6 \\ 4 \\ 6 \end{pmatrix} - \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}$

3. Find a normal vector to the plane by calculating the cross product of the two vectors.

$\mathbf{n} = \vec{PR} \times \vec{PQ} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix} \times \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} = \begin{pmatrix} (2)(6) - (0)(4) \\ (0)(2) - (2)(6) \\ (2)(4) - (2)(2) \end{pmatrix} = \begin{pmatrix} 12 \\ -12 \\ 4 \end{pmatrix}$
To simplify, we can divide by 4 to get a simpler normal vector: $\begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix}$.

4. Form the cartesian equation of the plane.

The equation of a plane is $\mathbf{r} \cdot \mathbf{n} = \mathbf{p} \cdot \mathbf{n}$, where $\mathbf{p}$ is a point on the plane.
Using point $P(4, 0, 0)$:
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix} = (4)(3) + (0)(-3) + (0)(1) = 12$
Writing $\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, we obtain:
$3x - 3y + z = 12$ (Shown)

1. Identify the coordinates or position vector of point $C$.

Since $OABC$ is a rectangle with $AB = 4$, $OC = 4$. Because $\mathbf{j}$ is a unit vector in the direction of $OC$, the position vector of $C$ is $\mathbf{c} = \begin{pmatrix} 0 \\ 4 \\ 0 \end{pmatrix}$.

2. Apply the perpendicular distance formula from a point to a plane.

The equation of the plane $PQR$ is $3x - 3y + z = 12$, which gives a normal vector $\mathbf{n} = \begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix}$.

The perpendicular distance from $C$ to the plane is given by $D = \frac{|Ax_1 + By_1 + Cz_1 - d|}{\sqrt{A^2 + B^2 + C^2}}$:

$\text{Distance} = \frac{|3(0) - 3(4) + 1(0) - 12|}{\sqrt{3^2 + (-3)^2 + 1^2}}$
$\text{Distance} = \frac{|-12 - 12|}{\sqrt{9 + 9 + 1}} = \frac{24}{\sqrt{19}} \approx 5.51 \text{ cm}$

1. Identify the normal vectors of the planes $PQR$ and $OABC$.

From part (i), the equation of plane $PQR$ is $3x - 3y + z = 12$, so its normal vector is $\mathbf{n}_1 = \begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix}$.
The plane $OABC$ is the $xy$-plane, so its normal vector is $\mathbf{n}_2 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

2. Calculate the acute angle between the two planes.

Let $\theta$ be the acute angle between the planes.
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$
$\cos \theta = \frac{\left| \begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right|}{\sqrt{3^2 + (-3)^2 + 1^2} \sqrt{0^2 + 0^2 + 1^2}}$
$\cos \theta = \frac{1}{\sqrt{19}(1)}$
$\theta = \cos^{-1}\left(\frac{1}{\sqrt{19}}\right) \approx 76.736^\circ$
$\theta \approx 76.7^\circ \text{ (1 d.p.)}$

1. Find the position vectors of $P$ and $Q$.

Given $OA = 6$ and $OP : PA = 2 : 1$, we have $OP = 4$.
$\vec{OP} = \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix}$
Point $Q$ lies on $BF$, which is vertical with height $10$. Given $BQ : QF = 3 : 2$, $BQ = 6$.
The position of $B$ is $6\mathbf{i} + 4\mathbf{j}$.
$\vec{OQ} = \begin{pmatrix} 6 \\ 4 \\ 6 \end{pmatrix}$

2. Form the vector equation of the line $PQ$.

The direction vector of $PQ$ is:
$\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 6 \\ 4 \\ 6 \end{pmatrix} - \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}$
The equation of line $PQ$ is:
$\mathbf{r} = \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 + 2\lambda \\ 4\lambda \\ 6\lambda \end{pmatrix}$

3. Find the intersection point $M$ with the plane $OCGD$.

The plane $OCGD$ contains the $y$-axis (through $C$) and $z$-axis (through $D$). Its equation is $x = 0$.
Substituting the $x$-component of the line $PQ$ into the plane equation:
$4 + 2\lambda = 0 \implies \lambda = -2$
Substituting $\lambda = -2$ into the line equation:
$\vec{OM} = \begin{pmatrix} 4 + 2(-2) \\ 4(-2) \\ 6(-2) \end{pmatrix} = \begin{pmatrix} 0 \\ -8 \\ -12 \end{pmatrix}$
Thus, the position vector of $M$ is $-8\mathbf{j} - 12\mathbf{k}$.

4. Deduce the distance from $M$ to the plane $OABC$.

The plane $OABC$ is the $xy$-plane, which has equation $z = 0$.
The distance from $M(0, -8, -12)$ to the $xy$-plane is simply the magnitude of its $z$-coordinate:
$\text{Distance} = |-12| = 12\text{ cm}$

1. Determine the coordinates of points $P$, $Q$, and $R$ and find their reflections about the plane $OABC$.

The plane $OABC$ is the $xy$-plane (where $z = 0$).
From the given geometry, the coordinates of the points in the original plane $PQR$ are:
$P(4, 0, 0)$
$Q(6, 4, 6)$
$R(6, 2, 0)$

Reflecting these points about the $xy$-plane negates their $z$-coordinates. The reflected points $P'$, $Q'$, and $R'$ are:
$P'(4, 0, 0)$
$Q'(6, 4, -6)$
$R'(6, 2, 0)$

2. Calculate two direction vectors for the reflected plane.

We use the vectors $\vec{P'R'}$ and $\vec{P'Q'}$ as our direction vectors:
$\vec{P'R'} = \begin{pmatrix} 6 \\ 2 \\ 0 \end{pmatrix} - \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix} = 2\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$

$\vec{P'Q'} = \begin{pmatrix} 6 \\ 4 \\ -6 \end{pmatrix} - \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -6 \end{pmatrix} = 2\begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$

3. Write the vector equation of the reflected plane.

Using $P'(4, 0, 0)$ as the position vector and the simplified direction vectors:
$\mathbf{r} = \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$
(Note: Other valid point/direction vector combinations are acceptable.)