We are given the point $P = (6, 1, -10)$ and the vector $\mathbf{v} = [-16, -8, 5]$. The arrow $\overrightarrow{PQ}$ represents the vector $\mathbf{v}$, which means:

$\overrightarrow{PQ} = \mathbf{v} = [-16, -8, 5]$

To find the vector $\overrightarrow{OQ}$, we use the fact that $Q$ can be determined by adding the components of vector $\mathbf{v}$ to the coordinates of point $P$.

So, if $P = (6, 1, -10)$ and $\mathbf{v} = [-16, -8, 5]$, then the coordinates of point $Q$ will be:

$Q = P + \mathbf{v} = (6, 1, -10) + (-16, -8, 5)$

Let's calculate the coordinates of $Q$:

$Q_x = 6 + (-16) = -10$ $Q_y = 1 + (-8) = -7$ $Q_z = -10 + 5 = -5$

Thus, $Q = (-10, -7, -5)$.

The vector $\overrightarrow{OQ}$ is simply the position vector of $Q$ relative to the origin $O(0, 0, 0)$, which is the same as the coordinates of $Q$:

$\overrightarrow{OQ} = [-10, -7, -5]$

So, the vector $\overrightarrow{OQ}$ is:

$\boxed{[-10, -7, -5]}$

Would you like more details on vector operations or any clarification?

Here are 5 related questions for further exploration:

1. How do you calculate the magnitude of a vector in $\mathbb{R}^3$?
2. What is the dot product of two vectors in $\mathbb{R}^3$, and how is it used?
3. How do you find the cross product of two vectors in $\mathbb{R}^3$?
4. How do you determine if two vectors are perpendicular in $\mathbb{R}^3$?
5. Can you explain how to compute the projection of one vector onto another in $\mathbb{R}^3$?

Tip: When adding vectors, always apply the operation component-wise: add corresponding $x$, $y$, and $z$-coordinates.