The matrix representation of a reflection operator across the plane $x+2y+3z=0$ Let $T:\mathbb{R}^3\rightarrow \mathbb{R}^3$ be the reflection across the plane $x+2y+3z=0$, find the matrix of this linear operator $T$ in respect to the basis $B=\left\{v_1,v_2,v_3\right\}$, where we have:
$v_1=\begin{bmatrix}1\\ 1\\ -1\end{bmatrix}$  $v_2=\begin{bmatrix}-1\\ 2\\ -1\end{bmatrix}$ $v_3=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$
First of what does it mean that $T$ is a reflection across the given plane? What will hapen to let's say the vector $(1,0,0)$ if it's passed through $T$? In my textbook, there was a short example that mentioned a reflection operator, which transformed a given vector $(a,b,c)$ to the vector $(a,b,-c)$ So that would be the reflection across the standard plane $x+y+z=0$, but here, will it act as if given a vector $(a,b,c)$ you get out a vector $(a,2b,-3c)$?
EDIT:
after reading the comments, I got the matrix representation as:
$T=\begin{bmatrix}1&-1&-1\\ 1&2&-2\\ -1&-1&-3\end{bmatrix}$
Now how can I find the representation of this operator in respect to the standard basis vectors $(e_1,e_2,e_3)$?
I read about finding the change of basis matrix so I calculated $T^{-1}$:
$T^{-1}=\begin{bmatrix}\frac{4}{7}&\frac{1}{7}&-\frac{2}{7}\\ -\frac{5}{14}&\frac{2}{7}&-\frac{1}{14}\\ -\frac{1}{14}&-\frac{1}{7}&-\frac{3}{14}\end{bmatrix}$
Is this it or is there more to it?

As noted above, vectors $v_1,v_2 \in p$ (plane),and $v_3$ is $\perp$ to $v_1,v_2$.
Since $v_1,v_2$ are in the plane their reflectection will be the same vector $v_1,v_2$ and since $v_3$ is the normal vector for this plane it's reflection will be the same vector just in the opposite direction ( $-v_3$).
So your reflection matrix in the base $B=\{v_1,v_2,v_3\}$ will look like this:
$R=\begin{bmatrix}
1 & 0 & 0\\ 
0 & 1 & 0\\ 
0 & 0 & -1
\end{bmatrix}_B$
To get the matrix representation in the standard base you can use the change of basis matrix $T$.
Notice that $T^{-1}$ is the matrix which will change $B_{std} \rightarrow B$ and $T$ will change $B\rightarrow B_{std}$.
The matrices will look like this:
$T^{-1}=\begin{bmatrix}
 \frac{4}{7}& \frac{1}{7} & \frac{-2}{7}\\ 
& & &\\
\frac{-5}{14} & \frac{2}{7} & \frac{-1}{14}\\ 
& & &\\
\frac{1}{14} & \frac{1}{7} & \frac{3}{14}
\end{bmatrix}
, T=\begin{bmatrix}
1 & -1 & 1\\ 
1 & 2 & 2 \\ 
-1 & -1 & 3 
\end{bmatrix}$
And finally $R'=TRT^{-1}=\begin{bmatrix}
 \frac{6}{7} & \frac{-2}{7} & \frac{-3}{7}\\ 
&&&\\
\frac{-2}{7} & \frac{3}{7} & \frac{-6}{7}\\ 
&&&\\
\frac{-3}{7} & \frac{-6}{7} & \frac{-2}{7}
\end{bmatrix}$
