A certain unique rotation matrix

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 ($p^2+3q^2=\cos^2(\theta)+\sin^2(\theta)=1$).

This most likely leads to the same set of solution as Yves Daoust's recipe. As I often use this method to generate examples of orthogonal matrices with rational entries (and it can be tweaked to satisfy the condition on coordinate sums as well) I will share it with you anyway.

Let $\vec{u}=(u_1,u_2,u_3)$ be any non-zero vector in $\Bbb{R}^3$. It is easy to see that the mapping $$ S:\vec{x}\mapsto\vec{x}-2\,\frac{\vec{x}\cdot\vec{u}}{\Vert\vec{u}\Vert^2}\,\vec{u} $$ is the orthogonal reflection with respect to the plane through the origin that has $\vec{u}$ as its normal. In particular $S$ is a length and angle preserving linear transformation. Therefore the triple $S(\mathbf{i})$, $S(\mathbf{j})$, $S(\mathbf{k})$ is an orthonormal system. Do observe that as a reflection $S$ changes the handedness, so the above system is left-handed. This poses no problem whatsoever, because rewriting them in the order $S(\mathbf{i})$, $S(\mathbf{k})$, $S(\mathbf{j})$ we get a right-handed orthonormal system. A $3\times3$ matrix with columns (or rows) forming a right-handed orthonormal system is a rotation matrix.

Observe that if we pose the extra constraint $u_1+u_2+u_3=0$, then from the formula for $S$ it is obvious that the coordinate sums of $\vec{x}$ and $S(\vec{x})$ are equal. Geometrically this follows from the fact that the reflected images differ from each other by a multiple of $\vec{u}$.

Let's see a few examples. The choice $\vec{u}=(1,1,-2)$ gives your example $$ \begin{aligned} S(\vec{i})&=(1,0,0)-\frac26\vec{u}&=(\frac23,-\frac13,\frac23),\\ S(\vec{k})&=(0,0,1)+\frac46\vec{u}&=(\frac23,\frac23,-\frac13),\\ S(\vec{j})&=(0,1,0)-\frac26\vec{u}&=(-\frac13,\frac23,\frac23). \end{aligned} $$ The choice $\vec{u}=(1,2,-3)$ gives another common example $$ \begin{aligned} S(\vec{i})&=(1,0,0)-\frac2{14}\vec{u}&=(\frac67,-\frac27,\frac37),\\ S(\vec{k})&=(0,0,1)+\frac6{14}\vec{u}&=(\frac37,\frac67,-\frac27),\\ S(\vec{j})&=(0,1,0)-\frac4{14}\vec{u}&=(-\frac27,\frac37,\frac67). \end{aligned} $$

Observe that by using a vector $\vec{u}$ with rational entries we get only rational coordinates. Because $\vec{u}$ and $\lambda\vec{u}$ determine the same reflection we might as well scale all the components of $\vec{u}$ to be integers.

It may be worth pointing out that swapping the order of $S(\mathbf{j})$ and $S(\mathbf{k})$ can be viewed as another reflection (w.r.t. to the plane bisecting the angle between those two vectors and containing $S(\vec{i})$. A composition of two reflections is a rotation.

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Adding an argument settling the auxiliary question about the sum of the coordinates.

If $M$ is an orthogonal real $3\times3$ matrix with constant row sum $\lambda$, then it follows that the vector $(1,1,1)^T$ is an eigenvector of $M$ belonging to eigenvalue $\lambda$. Because multiplication by $M$ preserves lengths of vectors, we must have $|\lambda|=1$. Because $\lambda$ is real we must have $\lambda=\pm1$. Both signs occur. For if $M$ is a rotation matrix with row sum $+1$, then the row sums of $-M$ are all $-1$. By swapping two rows of $-M$ it becomes an orthogonal matrix with row sum $=-1$.

Observe that when the angle of a 3D rotation is not an integer multiple of $180$ degrees, then $+1$ is the only real eigenvalue of the rotation matrix (the other two being $e^{\pm i\theta}$ where $\theta$ is the angle of rotation). This also shows in Yves' answer in his observation that the rotation must have $(1,1,1)$ as its axis.