Determining the transformation matrix R Find the transformation matrix R that describes a rotation of $120$ degrees about an axis from the origin through the point $(1,1,1)$. The rotation is clockwise as you look down the axis towards the origin.
It matters not which axis about which I wish for the rotation to occur. Let's suppose the rotation of the coordinate system is about the z-axis.
This means only the x and y axis will be rotating clockwise.
Let the rotated system be the $\bar{x}$ and $\bar{y}$ axis.
Let $A$ be the vector through $(1,1,1)$, 
$A_{x}=A\cos\theta$ and $A_{y}=A\sin\theta$
I've drawn diagrams but unsure how to proceed. Any help is appreciated.
 A: Perhaps I’ve misunderstood the problem, but it seems pretty straightforward to me.  
You’re being asked to express a $120$-degree clockwise rotation about the line through $(1,1,1)$ and the origin. If you sight back along this line towards the origin, the coordinate axes (i.e., their projections onto the plane through the origin and normal to the vector $\langle1,1,1\rangle^T$) are evenly spaced. So, a $120$-degree rotation will simply permute the coordinate axes. Remembering that the columns of a transformation matrix are the images of the basis vectors, we can immediately write down the matrix for this rotation:$$R=\pmatrix{0&1&0\\0&0&1\\1&0&0}.$$  
As a check, the eigenvalues of $R$ are $1$ and $-\frac12\pm i\frac{\sqrt3}2$ (i.e., the cube roots of unity), which indeed corresponds to a $120$-degree rotation. $\langle1,1,1\rangle^T$ is an eigenvector of $1$, so we have the correct axis, too.
A: Rotation about any freely chosen  axis in 3 dimensional space you can calculate using Rodrigues Formula 
$R=I+\sin(\theta)S(v)+(1-\cos(\theta))S^2(v)$
where $S(v)$ is  a skew-symmetric matrix assigned to the vector $v$ 
$S(v)=\begin{bmatrix}
0 & -v_z & v_y  \\
v_z & 0 & -v_x \\
-v_y & v_x & 0
\end{bmatrix}$
The only prerequisite for using this formula is to have the axis represented by the unit vector - in your case it would be the vector $v= \left[ \dfrac {1} {\sqrt{3} }  \  \dfrac{1}{\sqrt{3}} \ \  \dfrac {1}  {\sqrt{3}} \right]^T$.   
As you wish to have clockwise rotation about this axis you should assign   $\theta \ = 
-120^{\circ}$, minus because according to the right hand rule the positive angle is assigned to anticlockwise rotation.
