Assuming $av_1+bv_2+cv_3=v$, we have $$\begin{cases} b+2c=0\\ -2a+2b=0\\ 2a-c=5 \end{cases}$$ and by solving the sysytem, you have got $a=2$, $b=2$, and $c=-1$.
To find $Av$, as you mentioned, we have $$Av=A(2v_1+2v_2-v_3)=-2v_1+2v_2-4v_3= \begin{bmatrix}-6\\8\\0\end{bmatrix}$$
The $1$-eigenspace is the kernel of the map $I-A$, i.e., the null-space of the matrix
$$\begin{bmatrix}
0 & 0 & 0 \\
-1 & -2 & -1 \\
2 & 4 & 2
\end{bmatrix}.$$
If $(I-A)x=0$, then
$$\begin{bmatrix}
0 & 0 & 0 \\
-1 & -2 & -1 \\
2 & 4 & 2
\end{bmatrix}\begin{bmatrix}
x_1\\ x_2 \\ x_3
\end{bmatrix}=\begin{bmatrix}
0\\ -x_1-2x_2-x_3\\
2x_1+4x_2+2x_3
\end{bmatrix}=\begin{bmatrix}
0\\0\\0
\end{bmatrix}.$$
The only constraints are then $-x_1-2x_2-x_3=0$ and $2x_1+4x_2+2x_3=0$, which is really just the one equation $x_1+2x_2+x_3=0$. It helps to pick parameters $x_2=s$ and $x_3=t$ (these are the "free-variables"). Then the null-space consists of all $(x_1,x_2,x_3)$ satisfying
\begin{align}
x_1&=-2s-t\\
x_2&=s\\
x_3&=t
\end{align}
The null-space is thus
$$\operatorname{null}(I-A)\operatorname{span}\left\{\begin{bmatrix}-2\\ 1\\ 0 \end{bmatrix},\begin{bmatrix}-1\\ 0\\ 1 \end{bmatrix} \right\}$$
giving the two eigenvectors also verified by computer.
To find generalized eigenvectors, we need to find the null-space of $(I-A)^2$, but it turns out $(I-A)^2=0$. Thus
$$\operatorname{null}\left((I-A)^2\right)=\operatorname{span}\left\{\begin{bmatrix}1\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}0\\ 1\\ 0 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 1 \end{bmatrix} \right\}.$$
Conveniently, any vector not in the span of the two ordinary eigenvectors will thus work. Since those two lie in the plane $x_1+2x_2+x_3=0$, we can take the normal to this plane, i.e.,
$$\begin{bmatrix}
1\\2\\1
\end{bmatrix}.$$
Best Answer
What you are seeing is the intersection (meet) of two lines. Each line is defined from the connection (join) of two points. The $\times$ operator here isn't an actual cross product, because the cross product isn't defined for 4-vectors. I did not find where in the video they define the details of $\times$, but being an overview I doubt they went into this.
It is important to note that with planar homogeneous coordinates the line connecting two points is found using the vector cross product $\times$.
You can try it out, to find the $(a,b,c)$ coordinates of the line connecting the points $(x_1,y_1,1)$ and $(x_2,y_2,1)$ is
$$ \pmatrix{x_1 \\ x_1 \\ 1} \times \pmatrix{ x_2 \\ y_2 \\ 1} = \pmatrix{ y_1-y_2 \\ x_2 - x_1 \\ x_1 y_2 - x_2 y_1 }$$
making the equation of the line
$$ (y_1-y_2) x + (x_2-x_1) y + (x_1 y_2 - x_2 y_1) = 0 $$
Conversely, the point where two lines intersect is also found by the vector cross product $\times$. For example, two lines $u x + v y + w = 0$ and $p x + q y + r = 0$ meet at a point
$$ \pmatrix{ u \\ v \\ w} \times \pmatrix{p \\ q \\ r} = \pmatrix{ r v-q w \\ p w - r u \\ q u - p v } $$ with location $$ \pmatrix{x \\ y} = \pmatrix{ \frac{r v-q w}{q u - p v} \\ \frac{p w - r u}{q u - p v}}$$
So somehow in computer vision, the idea of some magical operator with symbol $\times$ exists that can be used to define the line where two points join ($v_1 \times v_2$) and the plane where two lines meet ($ (v_1 \times v_2) \times (v_2 \times v_3)$).
Note that the jump from 2D to 3D isn't trivial because of the way lines are defined.
In 2D we often use the $(a,b,c)$ coordinates for $ax + b y + c = 0$, where the vector $(a,b)$ is perpendicular to the line.
In 3D we often use Plücker coordinates $(\vec{e}, \vec{r}\times\vec{r})$ where the vector $\vec{e}$ is parallel to the line.
The realization I had is that a 2D line is a projection of a 3D plane that is perpendicular to the xy plane. A 3D plane is defined by the normal vector and the negative distance $(\vec{n},-d)$ and you get the exact 2D line coordinates if $\vec{n}=(a,b,0)$ and $d=-c$.
This is also evident that in 2D lines are dual to points, but in 3D lines are dual to lines, and points and dual to planes. So the $\times$ operator can be defined between dual objects (as well as the $\cdot$ operator for incidence) like points and lines in a plane, but not for points and lines in 3D.
So there is some trickery and magic going on here and they are glossing over some major details to get to a point.