Referred to the origin O, the position vectors of the points A and B are $i - j + 2k\quad\text{and}\quad 2i + 4j + k$ respectively.

1. Calculate the scalar (dot) product of the position vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$.

$\overrightarrow{OA} \cdot \overrightarrow{OB} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 4 \\ 1 \end{pmatrix}$
$= (1)(2) + (-1)(4) + (2)(1)$
$= 2 - 4 + 2$
$= 0$

2. State the conclusion based on the value of the scalar product.

Since $\overrightarrow{OA} \cdot \overrightarrow{OB} = 0$, $\overrightarrow{OA}$ is perpendicular to $\overrightarrow{OB}$ (shown).

1. Apply the ratio theorem formula for point M dividing the line segment AB in the ratio 1:2.

Since $AM : MB = 1 : 2$, using the ratio theorem:
$\overrightarrow{OM} = \frac{2\overrightarrow{OA} + 1\overrightarrow{OB}}{1 + 2}$

2. Substitute the given position vectors of A and B into the equation.

$\overrightarrow{OM} = \frac{2(\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + (2\mathbf{i} + 4\mathbf{j} + \mathbf{k})}{3}$

3. Expand and simplify to find the final position vector.

$\overrightarrow{OM} = \frac{1}{3} \left[ (2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}) + (2\mathbf{i} + 4\mathbf{j} + \mathbf{k}) \right]$
$\overrightarrow{OM} = \frac{1}{3} (4\mathbf{i} + 2\mathbf{j} + 5\mathbf{k})$
$\overrightarrow{OM} = \frac{4}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{5}{3}\mathbf{k}$

1. Calculate the vector product of $\overrightarrow{OA}$ and $\overrightarrow{OC}$.

$\overrightarrow{OA} \times \overrightarrow{OC} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} -4 \\ 2 \\ 2 \end{pmatrix}$

$= \begin{pmatrix} (-1)(2) - (2)(2) \\ (2)(-4) - (1)(2) \\ (1)(2) - (-1)(-4) \end{pmatrix}$

$= \begin{pmatrix} -6 \\ -10 \\ -2 \end{pmatrix}$

2. Calculate the magnitude of the resulting vector product.

$|\overrightarrow{OA} \times \overrightarrow{OC}| = \sqrt{(-6)^2 + (-10)^2 + (-2)^2}$
$= \sqrt{36 + 100 + 4}$
$= \sqrt{140}$
$= 2\sqrt{35}$

3. Calculate the exact area of triangle OAC using the formula.

$\text{Area} = \frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OC}|$
$= \frac{1}{2}(2\sqrt{35})$
$= \sqrt{35}$