\begin{equation}
Xa = \begin{bmatrix} a_1x_{1,1} + a_2x_{1,2}\\ a_1x_{2,1} + a_2x_{2,2}\\ 
 \end{bmatrix}
\end{equation}
and 
\begin{equation}
X^Tb = \begin{bmatrix} b_1x_{1,1} + b_2x_{2,1}\\ b_1x_{1,2} + b_2x_{1,2}\\ 
 \end{bmatrix}
\end{equation}
and 
\begin{equation}
Xa+X^Tb = \begin{bmatrix}a_1x_{1,1} + a_2x_{1,2} + b_1x_{1,1} + b_2x_{2,1}\\ a_1x_{2,1} + a_2x_{2,2}+ b_1x_{1,2} + b_2x_{2,2}\\ 
 \end{bmatrix} = \begin{bmatrix}(a_1+b_1)x_{1,1} + a_2x_{1,2} + b_2x_{2,1}\\ (a_2+b_2)x_{2,2} + a_1x_{2,1} + b_1x_{1,2}\\ 
 \end{bmatrix} = \begin{bmatrix}c_1 \\ c_2\\ 
 \end{bmatrix} 
\end{equation}
but $X$ is a rotation matrix gives us that 
\begin{equation}
Xa+X^Tb = \begin{bmatrix}(a_1+b_1)\cos\theta - a_2\sin\theta  + b_2\sin \theta \\ (a_2+b_2)\cos\theta  + a_1\sin\theta  - b_1\sin\theta\\ 
 \end{bmatrix} =  \begin{bmatrix}(a_1+b_1)\cos\theta +(b_2- a_2)\sin\theta  \\ (a_2+b_2)\cos\theta  + (a_1  - b_1)\sin\theta\\ 
 \end{bmatrix} 
\end{equation}
(if you only have $X^TX= I$ then you have to consider the extra case there $\sin \theta  \to -\sin \theta $; *i.e.* rotation composed with reflection. *Notice that $X$ is a rotation implies* $X^TX= I$!) and therefore that 
\begin{equation}
Xa+X^Tb =    \begin{bmatrix}(a_1+b_1)\cos\theta +(b_2- a_2)\sin\theta  \\ (a_2+b_2)\cos\theta  + (a_1  - b_1)\sin\theta\\ 
 \end{bmatrix} = \begin{bmatrix}c_1 \\ c_2\\ 
 \end{bmatrix} 
\end{equation}
and therefore (by using the triangle inequality) you don't have a solution if for example 
\begin{equation}
\frac{|(a_1+b_1)|}{\sqrt{2}}  +\frac{|(b_2- a_2)|}{\sqrt{2}} < |c_1|
\end{equation}
 and like wise for the second condition 
\begin{equation}
\frac{|(a_2+b_2)|}{\sqrt{2}}  +\frac{|(a_1- b_1)|}{\sqrt{2}} < |c_2|
\end{equation}
and you can probably come up with all kinds of other tests for failure, but here is the most general one:

An alternative/equivalent way to look at it is that you have an overdetermined system of 3 equations and 2 unknowns of the form 

\begin{equation}
\begin{array}
& ax & + & by & =& c_1\\
 cx & + & dy & =& c_2 \\
 x^2 & + & y^2 & =& 1 \\
\end{array}
\end{equation}

where $a = a_1+b_1$ , $b=a_2- b_2$, $c = a_2+b_2$, and $d = a_1- b_1$; which is highly unlikely to have solutions. 

> **Therefore you have solutions iff the solution to the system of
 equations**  \begin{equation} \begin{array} & ax & + & by & =& c_1 \\  cx
 & + & dy & =& c_2 \\ \end{array} \end{equation} **also satisfies the
> condition** $ x^2  +  y^2  = 1 $.

If you want to work out the $n=3$ case you can do the same exact thing but use the [Euler angles][1]; it will be long and tedious but you can probably get some kind of condition on the solutions. 


  [1]: https://en.wikipedia.org/wiki/Euler_angles