The relation between axes of 3D rotations Let's suppose we have two rotations about two different axes represented by vectors $v_1$ and $v_2$:
$R_1(v_1, \theta_1)$, $R_2(v_2,\theta_2)$. 
It's relatively easy to prove that composition of these two rotations gives rotation about axis $v_3$ distinct from axes $v_1$ and $v_2$ .   
Indeed
if for example $v_3=v_1$ then
$R_1(v_1, \theta_1) R_2(v_2,\theta_2)=R_3(v_1,\theta_3)$ leads to $R_2(v_2,\theta_2)=R_1^T(v_1, \theta_1)R_3(v_1,\theta_3)=R(v_1,\theta_3 -\theta_1)$ what gives $v_1=v_2$. ... Contradiction...
We see that composition of two rotations about different axes always generates a new axis of rotation.
The problem can be extended for condition of the plane generated by the axes.

Question: 
  
  
*
  
*Is it true that composition of two rotations generates the axis which
  doesn't belong to the plane   which is constructed by the original axes of rotations ?
  
*How to prove it ?
  
*If the statement is not however true what are conditions for not changing a plane during the composition of rotations $ ^{[1]}$ ?

$ ^{[1]}$ It can be observed that even in the case of quite regular rotations the above statement is true 
Let's take
$Rot(z,\dfrac{\pi}{2})Rot(x,\dfrac{\pi}{2})= \begin{bmatrix}
0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix} \begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 0 \\
\end{bmatrix} = \begin{bmatrix}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
\end{bmatrix} = Rot([1,1,1]^T, \dfrac{2}{3}\pi)$  
or
$Rot(x,  \pi  )Rot(z,  \pi  )= \begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{bmatrix} \begin{bmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix} = \begin{bmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1 \\
\end{bmatrix} = Rot( y, \pi)$  
So I suppose it is generally true but how to prove it ?

Using quaternions, as in Jyrki’s answer, is straightforward and provides some useful insights into the nature of the resultant rotation. As he suspects, the result can also be found another way.  
Suppose that the new rotation axis is coplanar with the original axes. Since vectors along the axis of rotation are eigenvectors of $1$, we must have $$R_1R_2(av_1+bv_2)=av_1+bv_2$$ for some $a$ and $b$ not both zero. Rearranging and again using the fact that $v_1$ and $v_2$ are eigenvectors of the respective rotations gives $$a(R_2-I)v_1=b(R_1^{-1}-I)v_2.$$ Wlog we can take $v_1$ and $v_2$ to be unit vectors. Substituting Rodrigues’ formula for the rotations then makes this $$a[(1-\cos{\theta_2})v_2\times(v_2\times v_1)+\sin{\theta_2}(v_2\times v_1)] = b[(1-\cos{\theta_1})v_1\times(v_1\times v_2)-\sin{\theta_1}(v_1\times v_2)]$$ which can be rearranged as $$[a(1-\cos{\theta_2})v_2+b(1-\cos{\theta_1})v_1]\times(v_2\times v_1) + (a\sin{\theta_2}-b\sin{\theta_1})(v_2\times v_1) = 0.\tag{*}$$ This clearly holds if $v_1$ and $v_2$ are colinear. If they aren’t, then the two vectors being added together in (*) are orthogonal, so they must both be zero. Any linear combination of $v_1$ and $v_2$ is orthogonal to $v_1\times v_2$, so the expression in square brackets must be zero, and since by hypothesis $v_1$ and $v_2$ are not colinear, we have $a(1-\cos{\theta_2})=0$ and $b(1-\cos{\theta_1})=0$, therefore at least one of the rotation angles must be zero.
