Relative to the origin $O$, two points $A$ and $B$ have position vectors given by $\mathbf{a} = 4\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$ and $\mathbf{b} = 4\mathbf{i} - 2\mathbf{j} + 7\mathbf{k}$ respectively. The point $P$ divides $AB$ in the ratio 3 : 1.

1. Apply the ratio theorem to find the position vector of $P$.

Since $P$ divides $AB$ in the ratio $3:1$, by the ratio theorem:
$\overrightarrow{OP} = \frac{1\mathbf{a} + 3\mathbf{b}}{3 + 1}$

2. Substitute the given vectors and simplify to find $\overrightarrow{OP}$.

$\overrightarrow{OP} = \frac{1}{4} \left( \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix} + 3 \begin{pmatrix} 4 \\ -2 \\ 7 \end{pmatrix} \right)$
$\overrightarrow{OP} = \frac{1}{4} \left( \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 12 \\ -6 \\ 21 \end{pmatrix} \right)$
$\overrightarrow{OP} = \frac{1}{4} \begin{pmatrix} 16 \\ -4 \\ 24 \end{pmatrix}$
$\overrightarrow{OP} = \begin{pmatrix} 4 \\ -1 \\ 6 \end{pmatrix}$

1. Calculate the position vector of $P$, $\overrightarrow{OP}$, based on the given ratio (as required from part (i)).

Using the ratio theorem for $P$ dividing $AB$ in the ratio $3 : 1$:
$\overrightarrow{OP} = \frac{1 \cdot \mathbf{a} + 3 \cdot \mathbf{b}}{3 + 1}$
$\overrightarrow{OP} = \frac{1}{4} \left( \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix} + 3 \begin{pmatrix} 4 \\ -2 \\ 7 \end{pmatrix} \right) = \frac{1}{4} \begin{pmatrix} 16 \\ -4 \\ 24 \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \\ 6 \end{pmatrix}$

2. Find the unit vector $\mathbf{c}$ in the direction of $\overrightarrow{OP}$.

The magnitude of $\overrightarrow{OP}$ is:
$|\overrightarrow{OP}| = \sqrt{4^2 + (-1)^2 + 6^2} = \sqrt{16 + 1 + 36} = \sqrt{53}$
Since $\mathbf{c}$ is the unit vector, we divide $\overrightarrow{OP}$ by its magnitude and express it as a column vector:
$\mathbf{c} = \frac{1}{\sqrt{53}} \begin{pmatrix} 4 \\ -1 \\ 6 \end{pmatrix}$

3. State the geometrical meaning of $|\mathbf{a} \cdot \mathbf{c}|$.

The scalar product of a vector $\mathbf{a}$ and a unit vector $\mathbf{c}$ gives the scalar projection of $\mathbf{a}$ onto the direction of $\mathbf{c}$. The modulus ensures the length is positive regardless of angle orientation.
Therefore, $|\mathbf{a} \cdot \mathbf{c}|$ represents the length of the projection of $\overrightarrow{OA}$ onto the line $OP$.

1. Determine the vectors $\overrightarrow{AO}$ and $\overrightarrow{AB}$.

$\overrightarrow{AO} = -\mathbf{a} = \begin{pmatrix} -4 \\ -2 \\ -3 \end{pmatrix}$
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 4 \\ -2 \\ 7 \end{pmatrix} - \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ -4 \\ 4 \end{pmatrix}$

2. Calculate the scalar product and magnitudes of $\overrightarrow{AO}$ and $\overrightarrow{AB}$.

$\overrightarrow{AO} \cdot \overrightarrow{AB} = (-4)(0) + (-2)(-4) + (-3)(4) = 0 + 8 - 12 = -4$
$|\overrightarrow{AO}| = \sqrt{(-4)^2 + (-2)^2 + (-3)^2} = \sqrt{16 + 4 + 9} = \sqrt{29}$
$|\overrightarrow{AB}| = \sqrt{0^2 + (-4)^2 + 4^2} = \sqrt{0 + 16 + 16} = \sqrt{32}$

3. Use the scalar product formula to find the angle $OAB$ to the nearest degree.

$\cos \angle OAB = \frac{\overrightarrow{AO} \cdot \overrightarrow{AB}}{|\overrightarrow{AO}||\overrightarrow{AB}|}$
$\cos \angle OAB = \frac{-4}{\sqrt{29}\sqrt{32}} = \frac{-4}{\sqrt{928}}$
$\angle OAB = \cos^{-1}\left(\frac{-4}{\sqrt{928}}\right) \approx 97.545^\circ$
Rounding to the nearest degree, $\angle OAB = 98^\circ$.