Finals studying continued.
$ T: \mathbb R^3{\rightarrow} \mathbb R^3$ with matrix
$$A= \begin{pmatrix} -2/7 & 6/7 & 3/7 \\ 3/7 & -2/7 & 6/7 \\ 6/7 & 3/7 & -2/7 \\ \end{pmatrix} $$ relative to the standard ordered basis for $\mathbb R^3$ is a rotation about some line through the origin $ (0,0,0) $.
I figure that since it is a symmetric matrix, $A $is invertible and $Q^{-1}AQ$ is diagonalizable. Also, $Q^{-1} = Q^{t} $ so I was thinking of finding the inverse of $A$. Or maybe try to find a vector that spans the line. I don't know.
EDIT: $A$ is not symmetric, so scratch that.
Edit:
As Lubin points out in his/her comment, your matrix $A$ does not represent a rotation. One can easily see that $AA^T\ne I$ and $\det(A)\ne1$. The $(2,2)$-th entry is probably wrong and it may be $-2/7$.If this is really the case,it's not hard to see that all row sums of the corrected $A$ are equal to $1$. Therefore the axis of rotation is the line spanned by $x=(1,1,1)^T$ (because $Ax=x$).In general, for a non-diagonal rotation matrix $A$, you can read off the axis from the skew-symmetric part of $A$ directly. See my answer to q766565 "Find the axis of rotation of a rotation matrix by INSPECTION (NOT by solving $Kv=v$)". In your example, the skew-symmetric part (up to a factor) of $A$ is equal to $$ W=A-A^T=\frac37\pmatrix{0&1&-1\\ -1&0&1\\ 1&-1&0}, $$ therefore the rotation axis is the span of $(w_{23},w_{31},w_{12})^T$, which is the line spanned by $(1,1,1)^T$.