Symmetry about the plane in affine space Let $H \subset \Bbb{R^3}$ be affine subspace defined by following equation:
$$x_1+2x_2-x_3=2$$
and $L$ is a line in $\Bbb{R^3}$ with parametric equation:
$$(0,1,1)+t(2,0,1)$$
Let $f:\Bbb{R^3}\to\Bbb{R^3}$ be a symmetry of the L about the $H$
Find the equation of $f(H_1)$, where $H_1$ is a plane also being set of solutions of following equation: $2x_1-x_2+x_3=1$
I don't even know where to start or what should I start with.
Any help and hints will be appreciated.
 A: HINT
To construct the matrix for $f$ let proceed as follow


*

*Translate $H\to H_0$ through the origin by $v_0=\left(-\frac13,-\frac23,\frac13\right)$


$$H_0: x_1+2x_2-x_3=0$$


*Consider a basis with $v_1,v_2 \in H_0$ and $v_3\parallel L$ that is


$$v_1=(1,0,1)\quad v_2=(2,-1,0)\quad v_3=(2,0,1)$$


*The symmetry matrix in this basis is


$$S_0=\begin{bmatrix}1&0&0\\0&1&0\\0&0&-1\end{bmatrix}$$


*The matrix for the change of basis form the given basis to the standard basis is


$$M=[v_1\,v_2\,v_3]=\begin{bmatrix}1&2&2\\0&-1&0\\1&0&1\end{bmatrix}$$


*The symmetry matrix in the standard basis is


$$R_0=MS_0M^{-1}$$


*Then $w=f(v)$ is represented by


$$w=R_0v-v_0$$


*Find a basis for $H_1$


$$w_1=(1,2,0)\quad w_2=(0,1,1)\quad w_0=\left(\frac26,-\frac16,\frac16\right)$$
$$P\in H_1: aw_1+bw_2+w_0$$


*Compute $f(w_1)$, $f(w_2)$ and $f(w_0)$ to find the image of $H_1$.

A: It’s still not entirely clear to me which of the possible reflections (symmetries) you’re meant to compute, so I’ll sketch out the overall construction methods.  
All of the symmetries leave the intersection of $L$ and $H$ fixed, so let’s start by computing that. I like working in homogeneous coordinates for problems like this. $L$ is the join of the points $\mathbf l_0 = [2:0:1:0]$ and $\mathbf l_1 = [0:1:1:1]$, and $\mathbf H = [1:2:-1:-2]$, so their intersection is $$\mathbf p = \mathbf l_0\mathbf l_1^T\mathbf H-\mathbf l_1\mathbf l_0^T\mathbf H = [-2:-1:-2:-1],$$ which corresponds to the inhomogeneous coordinates $\tilde{\mathbf p} = (2,1,2)^T$.  
The homogeneous vectors that represent planes are covariant: if we have a point transformation matrix $\mathbf x' = M\mathbf x$, then planes transform as $\mathbf\pi'=M^{-T}\mathbf\pi.$ We’re working with reflections, which are their own inverses, so once we have the point reflection matrix $R$, the reflection of the plane $\mathbf H_1$ Is $R^T\mathbf H_1$.  
The reflection will be a bit easier to construct if we decompose it as a translation of $\mathbf p$ to the origin, a reflection of some sort that leaves the origin fixed, and then a translation back, i.e., $R=T_{\mathbf p}R_OT_{-\mathbf p}$. This might seem like a lot of matrix multiplication at first glance, but it’s not. In block form, $$T_{\mathbf p}R_OT_{-\mathbf p} = \left[\begin{array}{c|c}I_3 & \tilde{\mathbf p} \\ \hline \mathbf 0^T & 1 \end{array}\right] \left[\begin{array}{c|c}R_3 & \mathbf 0 \\ \hline  \mathbf 0^T & 1 \end{array}\right] \left[\begin{array}{c|c}I_3 & -\tilde{\mathbf p} \\ \hline \mathbf 0^T & 1 \end{array}\right] = \left[\begin{array}{c|c}R_3 & (I_3-R_3)\tilde{\mathbf p} \\ \hline \mathbf 0^T & 1 \end{array}\right].$$ Here, $R_3$ is a $3\times3$ reflection matrix.  
All that’s left to do is to compute $R_3$, and that’s where we need to know exactly which reflection it is that is being computed. If it’s a reflection in the point $\mathbf p$, then $R_3=-I_3$. If it’s the usual orthogonal reflection in some plane, then you can use the standard formula $I-2{\tilde{\mathbf n}\tilde{\mathbf n}^T\over\tilde{\mathbf n}^T\tilde{\mathbf n}}$, where $\tilde{\mathbf n}$ is a (inhomogeneous) 3-d vector normal to the plane. If it’s the oblique reflection that gimusi suggests, then a way to construct it is to find a basis in which the matrix of the reflection is $\operatorname{diag}(-1,1,1)$ and perform a change of basis back to the standard basis. For the first vector in this basis, take $L$’s direction vector $(2,0,1)^T$. For the remaining two vectors, you need a linearly-independent pair of vectors parallel to $H$. $H$’s normal is $(1,2,-1)^T$. We can immediately read from it three vectors orthogonal to it: $(0,1,2)^T$, $(1,0,1)^T$ and $(2,-1,0)^T$. They’re obviously pairwise linearly independent, so any two of them will do. Taking the first two, we’d have $$R_3 = \begin{bmatrix}2&0&1\\0&1&0\\1&2&1\end{bmatrix} \begin{bmatrix} -1&0&0\\0&1&0\\0&0&1\end{bmatrix} \begin{bmatrix}2&0&1\\0&1&0\\1&2&1\end{bmatrix}^{-1}.$$ The resulting matrix $R_3$ will end up being the same regardless of the choice.
