First off, there is no simple rule (at least as far as I know) to represent a rotation as a cascade of single axis rotations. One mechanism is to compute a rotation matrix and then compute the single axis rotation angles. I should also mention that there is not a unique solution to the problem you are posing, since there are different cascades of rotations that will yield the same final position.
Let's first build two references frames, $A$ and $B$.
$A$ will denote the standard Euclidian basis. $B$ will denote the reference frame which results from applying the desired rotation. I will use subscript notation when referring to a vector in a specific reference frame, so $v_A$ is a vector whose coordinates are defined with respect to the standard Euclidian basis, and $v_B$ is defined w.r.t. $B$.
All we know so far about $B$ is that the vector $(0,0,1)_B = \frac{1}{\sqrt{77}}(8, -3, 2)_A$. We have one degree of freedom to select how $(0,1,0)_B$ is defined in reference frame $A$. We only require that it be orthogonal to $\frac{1}{\sqrt{77}}(8, -3, 2)_A$. So we would like to find a unit vector such that
$$8x-3y+2z=0$$
One solution is $\frac{1}{\sqrt{6}}(1,2,-1)_A$, which you can verify. This forces our hand for how the vector $(1,0,0)_B$ is represented w.r.t. $A$, hence
$$(1,0,0)_B=\frac{1}{\sqrt{462}}(-1,10,19)_A$$
So the rotation matrix from $B$ to $A$ is
$$R_B^A=\left( \begin{array}{ccc}
\frac{-1}{\sqrt{462}} & \frac{1}{\sqrt{6}} & \frac{8}{\sqrt{77}} \\
\frac{10}{\sqrt{462}} & \frac{2}{\sqrt{6}} & \frac{-3}{\sqrt{77}} \\
\frac{19}{\sqrt{462}} & \frac{-1}{\sqrt{6}} & \frac{2}{\sqrt{77}} \end{array} \right)$$
Given any vector with coordinates defined in the Euclidian basis (such as a point in the $xy$-plane), multiplying by the inverse this matrix will yield the representation in our new reference frame. If you want to search for the three axis rotations, I refer you to Wolfram MathWorld's page on Euler Angles.
Hope this helps! Let me know if anything is unclear or if I you think you see any mistakes.
Assuming that the vertices $A,B,C$ and $D$ are oriented as in your diagram, and $\theta$ measures the anticlockwise rotation:
Given $B'=(a',b')$ and $D'(c',d')$:
Translate coordinate system so that the rectangle's center is at the origin:
$$E' = (B' + D') / 2$$
$$B'_T = B' - E', \qquad D'_T = D' - E'$$
Rotate coordinate system so that the rectangle is oriented with its sides parallel to the $x,y$ axes (i.e. undo the rotation):
$$B_T=\left(\begin{matrix}
a_T \\
b_T\end{matrix}\right) = \left( \begin{matrix}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{matrix}\right)\left(\begin{matrix}
a'_T \\
b'_T\end{matrix}\right), \quad D_T=\left(\begin{matrix}
c_T \\
d_T\end{matrix}\right)=\left( \begin{matrix}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{matrix}\right)\left(\begin{matrix}
c'_T \\
d'_T\end{matrix}\right)$$
Find other corners $A_T,C_T$ by
$$A_T = \left(\begin{matrix}
a_T \\
d_T \end{matrix}\right), \qquad C_T = \left(\begin{matrix}
b_T \\
c_T \end{matrix}\right)$$
- Rotate $A_T$ and $C_T$ back to obtain $A'_T$ and $C'_T$:
$$A'_T=\left( \begin{matrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{matrix}\right)\left(\begin{matrix}
a_T \\
d_T\end{matrix}\right), \qquad C'_T=\left( \begin{matrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{matrix}\right)\left(\begin{matrix}
b_T \\
c_T\end{matrix}\right)$$
- Translate coordinate system back that the rectangle's center returns to where it was.
$$A' = A'_T + E', \qquad C' = C'_T + E'$$
Best Answer
The direction vector of the yellow line is
$$ \pmatrix{\cos\theta\cos\phi\\\cos\theta\sin\phi\\\sin\theta} $$
with $\theta=15^\circ$ and $\phi=60^\circ$. Thus its angle with the $x$ axis is
$$ \arccos\left(\cos\theta\cos\phi\right)=\arccos\left(\cos15^\circ\cos60^\circ\right)\approx61.1^\circ\;. $$
Since this is far from your measurement of $\approx28.2^\circ$, it seems that what you're interested in may in fact not be the angle of the yellow line with the $x$ axis. The image suggests that the angle you're measuring is the angle between the $x$ axis and the projection of the yellow line onto the $x$-$z$ plane. This is indeed
$$ \arctan\frac{\sin\theta}{\cos\theta\cos\phi}=\arctan\frac{\sin15^\circ}{\cos15^\circ\cos60^\circ}\approx28.18679^\circ\;. $$