A certain unique rotation matrix One can find that the matrix
$A=\begin{bmatrix}
    -\dfrac{1}{3} &     \dfrac{2}{3}      &     \dfrac{2}{3}   \\
     \dfrac{2}{3}  &    -\dfrac{1}{3}   &   \dfrac{2}{3}    \\
    \dfrac{2}{3} &    \dfrac{2}{3}  &    -\dfrac{1}{3}  \\
        \end{bmatrix}  $  
is at the same time $3D$ rotation  matrix and for it the sum of  entries  in  every  column (row)  is constant (here $-\dfrac{1}{3}+   \dfrac{2}{3}     +     \dfrac{2}{3} = 1)$.
 The same is true if we change  the order of columns in it.
For example:
$A_1=\begin{bmatrix}
    \dfrac{2}{3} &     \dfrac{2}{3}      &     -\dfrac{1}{3}   \\
     -\dfrac{1}{3}  &    \dfrac{2}{3}   &   \dfrac{2}{3}    \\
    \dfrac{2}{3} &    -\dfrac{1}{3}  &    \dfrac{2}{3}  \\
        \end{bmatrix}  $       $A_2=\begin{bmatrix}
     \dfrac{2}{3} &     -\dfrac{1}{3}      &     \dfrac{2}{3}   \\
     \dfrac{2}{3}  &     \dfrac{2}{3}   &   -\dfrac{1}{3}    \\
    -\dfrac{1}{3} &    \dfrac{2}{3}  &     \dfrac{2}{3}  \\
        \end{bmatrix}  $   
Questions:
Is it any systematic way to find other non-trivial (without $0$ and $1$) rotation matrices with this property?
Especially it is interesting whether the above rotation matrices are the only ones with rational entries ?- maybe someone knows other rotation matrices exist where the sum of entries is constant..
and...
Can it be proved in some way that if the sum of entries in columns for a  rotation matrix is constant then it should be equal to the length of  column vectors?

The three row vectors of the matrix define a triorthonormal frame, so that the tips of the vectors belong to a unit sphere. The sum of entries being a constant corresponds to a plane in the first octant, $x+y+z=c$, which intersects the sphere along a (small) circle.
You can take any three points on that circle that form an equilateral triangle and they fulfill the conditions on the rows (in addition to orthogonality). The transform is a rotation around the first octant bissector.
As the transform is orthogonal, the transpose of the matrix corresponds to the inverse rotation, also a rotation around the bissector, so that the condition on the rows also holds.
By the Rodrigues formula,
$$M=\cos(\theta)\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\right)+\frac{1-\cos(\theta)}3\left(\begin{matrix}1&1&1\\1&1&1\\1&1&1\end{matrix}\right)+\frac{\sin(\theta)}{\sqrt3}\left(\begin{matrix}\ \ 0&\ \ 1&-1\\-1&\ \ 0&\ \ 1\\\ \ 1&-1&\ \ 0\end{matrix}\right).$$
There is a simple infinity of solutions. Rational ones occur for $\sin(\theta)=0$ or $\sin(\theta)=\pm\frac{\sqrt3}2$, but this brings nothing new. Other ones are found when $\cos(\theta)$ and $\dfrac{\sin(\theta)}{\sqrt3}$
are simultaneously rational ($p^2+3q^2=\cos^2(\theta)+\sin^2(\theta)=1$).
