To build a Virtual Reality (VR) experience, a virtual world is created to contain geometric models. With reference to the origin $O$, every point in the virtual world is represented as coordinates $(x, y, z)$. Perspective projection is a technique used to project points in the virtual world onto a virtual screen while respecting the scaling properties of objects at various distances. Getting this right in VR helps in the perception of depth and scale. In addition, only points that lie within a zone in front of the eye positioned at $O$, known as the viewing frustum (see Fig. 1), will be projected onto the virtual screen. In Fig. 1, a triangular object $ABC$ in the viewing frustum with $A(36,36,12)$ and $B(24,48,72)$ is projected onto the virtual screen to give an image of $A'B'C'$. It is given that the coordinates of $A'$ are $(3,3,1)$ and the line that passes through $A'$ and $B'$ is parallel to the vector $2\mathbf{i}+\mathbf{j}-2\mathbf{k}$.

1. Form the vector equation of the line passing through the origin $O$ and point $B$.

Since the projection is from $O$, the image point $B'$ must lie on the line passing through $O(0, 0, 0)$ and $B(24, 48, 72)$.
The direction vector of this line is $\overrightarrow{OB} = \begin{pmatrix} 24 \\ 48 \\ 72 \end{pmatrix}$, which is parallel to $\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$.
The equation of line $OB$ is:
$l_{OB}: \mathbf{r} = \mu \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$

2. Form the vector equation of the line passing through $A'$ and $B'$.

We are given that $A'$ has coordinates $(3, 3, 1)$ and the line $A'B'$ is parallel to $2\mathbf{i} + \mathbf{j} - 2\mathbf{k}$.
The equation of line $A'B'$ is:
$l_{A'B'}: \mathbf{r} = \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}$

3. Equate the two line equations to find their point of intersection, $B'$.

Since $B'$ lies on both lines, we equate their position vectors:
$\begin{pmatrix} \mu \\ 2\mu \\ 3\mu \end{pmatrix} = \begin{pmatrix} 3 + 2\lambda \\ 3 + \lambda \\ 1 - 2\lambda \end{pmatrix}$
This gives a system of linear equations:
$\mu = 3 + 2\lambda \quad \text{--- (1)}$
$2\mu = 3 + \lambda \quad \text{--- (2)}$
$3\mu = 1 - 2\lambda \quad \text{--- (3)}$

4. Solve the system of equations for the parameters $\lambda$ and $\mu$.

Substitute equation (1) into equation (2):
$2(3 + 2\lambda) = 3 + \lambda$
$6 + 4\lambda = 3 + \lambda$
$3\lambda = -3 \implies \lambda = -1$
Substitute $\lambda = -1$ into equation (1):
$\mu = 3 + 2(-1) = 1$
(Check for consistency with equation (3): $3(1) = 3$ and $1 - 2(-1) = 3$. The lines intersect consistently.)

5. Determine the position vector and coordinates of $B'$.

Using $\mu = 1$, the position vector of $B'$ is:
$\overrightarrow{OB'} = 1 \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$
Hence, the coordinates of $B'$ are $(1, 2, 3)$.

1. Find two non-parallel direction vectors that lie on the virtual screen.

From part (i), the line through $A'$ and $B'$ is parallel to $\mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}$.

Using the coordinates of $A'$ and $C'$, we find another direction vector on the plane:
$\overrightarrow{A'C'} = \overrightarrow{OC'} - \overrightarrow{OA'} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$

2. Determine a normal vector to the virtual screen by taking the cross product of the two direction vectors.

$\mathbf{n} = \mathbf{d}_1 \times \overrightarrow{A'C'} = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} \times \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$
$\mathbf{n} = \begin{pmatrix} (1)(1) - (-2)(1) \\ (-2)(-2) - (2)(1) \\ (2)(1) - (1)(-2) \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix}$

3. Formulate the equation of the plane in scalar product form $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$.

$\mathbf{r} \cdot \mathbf{n} = \overrightarrow{OA'} \cdot \mathbf{n}$
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix} = 3(3) + 3(2) + 1(4)$
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix} = 19$

1. Find the displacement vectors $\overrightarrow{A'B'}$ and $\overrightarrow{A'C'}$.

$\overrightarrow{A'B'} = \overrightarrow{OB'} - \overrightarrow{OA'} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix}$
$\overrightarrow{A'C'} = \overrightarrow{OC'} - \overrightarrow{OA'} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$

2. Evaluate the cross product of the two displacement vectors.

$\overrightarrow{A'B'} \times \overrightarrow{A'C'} = \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} (-1)(1) - (2)(1) \\ (2)(-2) - (-2)(1) \\ (-2)(1) - (-1)(-2) \end{pmatrix} = \begin{pmatrix} -3 \\ -2 \\ -4 \end{pmatrix}$
(Note: This cross product may have already been calculated in part (ii) to find the normal of the plane.)

3. Calculate the area of the triangle using the magnitude of the cross product.

$\text{Area of } \triangle A'B'C' = \frac{1}{2} |\overrightarrow{A'B'} \times \overrightarrow{A'C'}|$
$\text{Area} = \frac{1}{2} \sqrt{(-3)^2 + (-2)^2 + (-4)^2}$
$\text{Area} = \frac{1}{2} \sqrt{9 + 4 + 16} = \frac{\sqrt{29}}{2} \text{ units}^2$

1. Express the position vector of $C$ in terms of an unknown scalar.

Since $C'$ is the perspective projection of $C$ from the projection centre $O$, the points $O$, $C'$, and $C$ must be collinear. Thus, $\overrightarrow{OC}$ is a scalar multiple of $\overrightarrow{OC'}$. Since $C$ lies in front of $O$, we have $\overrightarrow{OC} = k \overrightarrow{OC'}$ for some scalar $k > 0$.
$\overrightarrow{OC} = k \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} k \\ 4k \\ 2k \end{pmatrix}$

2. Formulate the vector $\overrightarrow{BC}$ in terms of $k$.

Using the position vector of $B$, $\overrightarrow{OB} = \begin{pmatrix} 24 \\ 48 \\ 72 \end{pmatrix}$:
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} k \\ 4k \\ 2k \end{pmatrix} - \begin{pmatrix} 24 \\ 48 \\ 72 \end{pmatrix} = \begin{pmatrix} k - 24 \\ 4k - 48 \\ 2k - 72 \end{pmatrix}$

3. Express the length of projection of $\overrightarrow{BC}$ onto the normal of the virtual screen.

From part (ii), the normal vector to the virtual screen is $\mathbf{n} = \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix}$. Its magnitude is $|\mathbf{n}| = \sqrt{3^2 + 2^2 + 4^2} = \sqrt{29}$.
The length of projection of $\overrightarrow{BC}$ onto the normal is given by $\frac{|\overrightarrow{BC} \cdot \mathbf{n}|}{|\mathbf{n}|}$:
$\overrightarrow{BC} \cdot \mathbf{n} = 3(k - 24) + 2(4k - 48) + 4(2k - 72)$
$= 3k - 72 + 8k - 96 + 8k - 288 = 19k - 456 = 19(k - 24)$
$\text{Length of projection} = \frac{|19(k - 24)|}{\sqrt{29}}$

4. Equate the projection length to the given value and solve for $k$.

$\frac{19|k - 24|}{\sqrt{29}} = \frac{57}{29}$
$|k - 24| = \frac{57}{29} \times \frac{\sqrt{29}}{19} = \frac{3\sqrt{29}}{29}$
$k = 24 \pm \frac{3\sqrt{29}}{29}$
Since both values of $k$ are positive, $C$ stays in front of the viewing frustum for both cases. Substituting these into $\overrightarrow{OC}$ yields the possible position vectors:
$\overrightarrow{OC} = \left( 24 + \frac{3\sqrt{29}}{29} \right) \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \quad \text{and} \quad \left( 24 - \frac{3\sqrt{29}}{29} \right) \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}$