Given the position vectors of the points $A$, $B$, and $C$:

$$\begin{aligned}\vec{OA} &= \hat{i} + 2\hat{j} - 3\hat{k} \\\vec{OB} &= 3\hat{i} - 4\hat{j} + 5\hat{k} \\\vec{OC} &= 5\hat{i} - 10\hat{j} + 13\hat{k}\end{aligned}$$

First, let us find the vector $\overrightarrow{AB}$:

$$\begin{aligned}\overrightarrow{AB} &= \vec{OB} - \vec{OA} \\&= (3\hat{i} - 4\hat{j} + 5\hat{k}) - (\hat{i} + 2\hat{j} - 3\hat{k}) \\&= (3 - 1)\hat{i} + (-4 - 2)\hat{j} + (5 - (-3))\hat{k} \\&= 2\hat{i} - 6\hat{j} + 8\hat{k}\end{aligned}$$

Next, let us find the vector $\overrightarrow{BC}$:

$$\begin{aligned}\overrightarrow{BC} &= \vec{OC} - \vec{OB} \\&= (5\hat{i} - 10\hat{j} + 13\hat{k}) - (3\hat{i} - 4\hat{j} + 5\hat{k}) \\&= (5 - 3)\hat{i} + (-10 - (-4))\hat{j} + (13 - 5)\hat{k} \\&= 2\hat{i} - 6\hat{j} + 8\hat{k}\end{aligned}$$

We can see that:

$$\overrightarrow{BC} = 1 \cdot \overrightarrow{AB}$$

Since $\overrightarrow{BC}$ is a scalar multiple of $\overrightarrow{AB}$ (i.e., $\overrightarrow{BC} = k\overrightarrow{AB}$ where $k = 1$), the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are parallel. Furthermore, since they share a common point $B$, the points $A$, $B$, and $C$ must lie on the same straight line. Therefore, the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are collinear.

Alternatively, we can show this by calculating the magnitudes of the vectors:

$$\begin{aligned}|\overrightarrow{AB}| &= \sqrt{2^2 + (-6)^2 + 8^2} \\&= \sqrt{4 + 36 + 64} \\&= \sqrt{104}\end{aligned}$$$$\begin{aligned}|\overrightarrow{BC}| &= \sqrt{2^2 + (-6)^2 + 8^2} \\&= \sqrt{4 + 36 + 64} \\&= \sqrt{104}\end{aligned}$$

Now, let us find the vector $\overrightarrow{AC}$:

$$\begin{aligned}\overrightarrow{AC} &= \vec{OC} - \vec{OA} \\&= (5\hat{i} - 10\hat{j} + 13\hat{k}) - (\hat{i} + 2\hat{j} - 3\hat{k}) \\&= (5 - 1)\hat{i} + (-10 - 2)\hat{j} + (13 - (-3))\hat{k} \\&= 4\hat{i} - 12\hat{j} + 16\hat{k}\end{aligned}$$

The magnitude of $\overrightarrow{AC}$ is:

$$\begin{aligned}|\overrightarrow{AC}| &= \sqrt{4^2 + (-12)^2 + 16^2} \\&= \sqrt{16 + 144 + 256} \\&= \sqrt{416} \\&= \sqrt{4 \times 104} \\&= 2\sqrt{104}\end{aligned}$$

Since,

$$\begin{aligned}|\overrightarrow{AB}| + |\overrightarrow{BC}| &= \sqrt{104} + \sqrt{104} \\&= 2\sqrt{104} \\&= |\overrightarrow{AC}|\end{aligned}$$

The sum of the lengths of $AB$ and $BC$ is equal to the length of $AC$. This confirms that the points $A$, $B$, and $C$ are collinear, and thus the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are collinear.