<s>[INST] Reflection across the plane Let $T: \Bbb R^3 \rightarrow \Bbb R^3$ be the linear transformation given by reflecting across the plane $S=\{x:-x_1+x_2+x_3=0\}$ (...) 
Then, $S={\rm gen}[(1,1,0),(1,0,1)].$
But how can I get the matrix $R_v$ such that reflects across $S$? 
Thanks!
 [/INST] The vector $\vec{n}=\langle -1,1,1 \rangle$ is normal to the plane, thus $T$ maps $\vec{n}$ to $-\vec{n}=\langle 1,-1,-1 \rangle$. It also fixes the vectors $\vec{u}= \langle 1,1, 0 \rangle$ and $\vec{v} = \langle -1,1,-2 \rangle$ that are perpendicular to each other and to $\vec{n}$.  More generaly, $\vec{u}$ could be any vector whose dot product with $\vec{n}$ is zero (such vectors are easy to find) and $\vec{v} = \vec{n}\times\vec{u}$.
In the basis determined by $\vec{n}$, $\vec{u}$, and $\vec{v}$, your matrix looks like
$$\left(
\begin{array}{ccc}
  -1 & 0 & 0 \\
  0 & 1  & 0 \\
  0 & 0 &  1 \\
\end{array}
\right).$$
The matrix $A$ whose columns are formed by $\vec{n}$, $\vec{u}$, and $\vec{v}$ form a similarity transformation that changes the diagonal matrix $E$ formed by the eigenvalues $-1$, $1$, and $1$ into the matrix you want.  Thus, your matrix is
$$A \cdot E \cdot A^{-1} = 
\left(
\begin{array}{ccc}
 -1 & 1 & -1 \\
 1 & 1 & 1 \\
 1 & 0 & -2 \\
\end{array}
\right)
\cdot
\left(
\begin{array}{ccc}
  -1 & 0 & 0 \\
  0 & 1  & 0 \\
  0 & 0 &  1 \\
\end{array}
\right)
\cdot
\frac{1}{6}\left(
\begin{array}{ccc}
 -2 & 2 & 2 \\
 3 & 3 & 0 \\
 -1 & 1 & -2 \\
\end{array}
\right)
=
\frac{1}{3}\left(
\begin{array}{ccc}
 1 & 2 & 2 \\
 2 & 1 & -2 \\
 2 & -2 & 1 \\
\end{array}
\right).$$
Of course, it's easy to check that this matrix behaves as advertised by multiplying it by the vectors above.
Addendum
This technique can be used to derive Bubba's answer, although I used Mathematica to do the heavy lifting for me.
A = Transpose[{{a, b, c}, {0, -c, b}, {b^2 + c^2, -a*b, -a*c}}];
FullSimplify[A.{{-1, 0, 0}, {0, 1, 0}, {0, 0, 1}}.Inverse[A]] /.
  a^2 + b^2 + c^2 -> 1 // MatrixForm

$$
\left(
\begin{array}{ccc}
 1-2 a^2 & -2 a b & -2 a c \\
 -2 a b & 1-2 b^2 & -2 b c \\
 -2 a c & -2 b c & 2 \left(a^2+b^2\right)-1 \\
\end{array}
\right)
$$
However, seeing Bubba's answer reminded me wonderful and simple Householder reflections are. Just define the function $f:{\mathbb R}^n \rightarrow {\mathbb R}^n$ by $f(x)=\vec{x}-2\vec{n}(\vec{n}^T\vec{x})$, where $\vec{x}$ is a column vector and $\vec{n}$ is a unit column vector denoting your normal.  It's super simple to check that $f$ maps $\vec{n}$ to $-\vec{n}$:
$$f(\vec{n}) = \vec{n}-2\vec{n}(\vec{n}^T\vec{n}) = \vec{n}-2\vec{n}1=-\vec{n},$$
where $\vec{n}^T\vec{n}=1$, since $\vec{n}$ is a unit vector.  It's also super simple to see that any vector $\vec{x}$ perpendicular to $\vec{n}$ is fixed by $f$, since
$$f(\vec{x}) = \vec{x}-2\vec{n}(\vec{n}^T\vec{x}) = \vec{x}-2\vec{n}0=\vec{x}.$$
Factoring out the $\vec{x}$ from the definition of $f$, we see the matrix you need:
$$I-2\vec{n}\,\vec{n}^T.$$
Applying this to your situation, we get the same matrix as above.
\begin{align}
I-\vec{n}\,\vec{n}^T &=
\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right)
-
2\frac{1}{\sqrt{3}}\left(
\begin{array}{c}
 -1 \\
 1 \\
 1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
 -1 & 1 &
   1 \\
\end{array}
\right)/\sqrt{3}\\
&= \left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right)
- 2\frac{1}{3}
\left(
\begin{array}{ccc}
 1 & -1 & -1 \\
 -1 & 1 & 1 \\
 -1 & 1 & 1 \\
\end{array}
\right)
=
\frac{1}{3}
\left(
\begin{array}{ccc}
 1 & 2 & 2 \\
 2 & 1 & -2 \\
 2 & -2 & 1 \\
\end{array}
\right)
\end{align}
</s>