Reflection xyz plane

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?

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 $$x - 3y + 2z = 0$$.