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I have this matrix:

$A=\dfrac17\begin{bmatrix}6 & 3 &-2\\3 & -2 & 6 \\ -2 & 6 & 3\end{bmatrix}$

We have that $A^TA=I$. Since $\det A=-1$, its is a reflection matrix.

My question is what is the reflecting plane?

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A key observation is that the normal vector $n$ of the reflecting plane gets sent to $-n$ by $A$. We find $An = -n = (-I)n$, where $I$ is the identity matrix, hence $(A + I)n = 0$.

Solving this for $n$ we find $n = (1, -3, 2)$ as a solution (up to a scalar multiple), so the reflecting plane is given by $x - 3y + 2z = 0$.

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