Take $\vec{B} = \begin{pmatrix}0 \\ 0\\ k\end{pmatrix} k>0$
Now, $\vec{F} = q\vec{v} \times \vec{B} = q\begin{pmatrix}3 \\-1\\ 0\end{pmatrix}\times\begin{pmatrix}0 \\ 0\\ k\end{pmatrix} = -q\begin{pmatrix}k \\ 3k\\ 0\end{pmatrix}$
$$\hat{F} = -q\frac{k}{|q|k\sqrt{1+9}}\begin{pmatrix}1 \\ 3\\ 0\end{pmatrix} = \frac{-\operatorname{sgn}(q)}{\sqrt{10}}\begin{pmatrix}1 \\ 3\\ 0\end{pmatrix}$$
Try taking dot products $\vec{F}\cdot\vec{v}$ and $\vec{F}\cdot\vec{B}$
I am doing some major guesswork here due to the text being quite confusing. (Where does $w$ come from? Why is $\{rv+sw+x|r,s\in\mathbb R\}$ being classified as a line, when it is clearly a plane?) So here is my attempt to clear your confusion.
Q1. Your solution is incorrect. You need to subtract $\bf x$ from the other vectors.
Q2. Here is also a point I'm confused about. The author is clearly talking about a line in that section of the text. Defining a line as the set $\{kv+x|k\in\mathbb R\}$, we see that we need a direction vector $v$ and a translation vector $x$. The translation vector $x$ defines a starting point of the line, and the direction vector $v$ extends the line in that direction (forwards and backwards). The subtraction $(rv+x)-(sv+x)=(r-s)v$ shows that the direction vector $v$ and the translation vector $x$ are independent, so $x$ can be chosen as any point on the line.
Q3. Subtracting two position vectors gives you the direction vector from one of the points to the other.
Now, to understand the construction of the plane $\{rv+sw+x|r,s\in\mathbb R\}$, we see that there are two direction vectors $v,w$ and a translation vector $x$. The direction vectors $v,w$ span our plane.
To see that $v,w$ are found by subtracting vectors, consider three given (non-collinear) points. They form a triangle. Choose one vertex as the starting point. We calculate the direction vectors from our starting point to the other two vertices. These two vectors are precisely $v,w$.
Finally, the plane $\mathbb R^3$ passing through the points $(1,2,3),(0,2,1),(3,-1,1)$ is given by $$\{r(-1,0,-2)+s(2,-3,-2)+(1,2,3)|r,s\in\mathbb R\}$$
Substituting $(r,s) = (0,0), (1,0),(0,1)$ shows that the plane above does pass through our points.
Best Answer
$\vec I$ represents true physical motion, so it must be reflected when you reflect the coordinates. We don't know about $\vec B$ until we chase through the situation.