What would be a suitable basis for:
The matrix $A$ which belongs to the reflection at the line spanned by $\begin{bmatrix} 2 \\ 2 \\ 4 \end{bmatrix}$
I simply don't really understand how a 3d reflection can be about a line. Would the basis just be a perpendicular vector to this line?
Note that reflection about $z$ axes can be defined simply by $$P(a,b,c) \to P'(-a,-b,c)$$
thus the trick here is to select a basis with $x\to v_1$, $y\to v_2$ and $z\to v_3=(2,2,4)$ since in this basis the reflection matrix with respect to "z axes", that is $v_3$ is
$$M=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&1 \end{bmatrix}$$
For $v_1$ and $v_2$ we van choose any 2 vectors orthogonal to $v_3$. Notably set $v_3=(1,1,2)$, $v_1=(1,1,-1)\perp v_3$ and for $v_2$ let consider
$$v_1\times v_3=\begin{bmatrix}i&j&k\\1&1&-1\\1&1&2 \end{bmatrix}=3i-3j$$
thus assume $v_2=(1,-1,0)$.
Then since the matrix of change of basis, from the new to the standard, is
$$N=\begin{bmatrix}1&1&1\\1&1&1\\-1&0&2 \end{bmatrix} \implies v_S=Nv_B$$
we have that the reflection matrix is
$$R=NMN^{-1}$$