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 ?

This is easily seen if we assume familiarity with the use of unit quaternions in representing rotations. A rotation $R$ about the axis given  $\vec{v}=v_1\bf{i}+v_2\bf{j}+v_3\bf{k}$ by the angle $\theta$ is represented by the quaternion
$$
q=\cos\frac\theta2+\sin\frac\theta2\vec{v}.
$$
Here it is essential that $\vec{v}$ is a unit vector.
The connection is that the rotated version $R\vec{u}$ of a vector $\vec{u}$ is then given by the quaternion product
$$
R\vec{u}=q\vec{u}\overline{q},
$$
where $\overline{q}=\cos\frac\theta2-\sin\frac\theta2\vec{v}$ is the conjugate quaternion.
The composition of two such rotations is then faithfully reproduced as a product of the representing quaternions. So if another rotation $R'$ is represented by $q'=\cos\frac\alpha2+\sin\frac\alpha2\vec{v}'$, the composition $R'\circ R$ is represented by the product
$$
\begin{aligned}
qq'&=\left(\cos\frac\alpha2\cos\frac\theta2-\sin\frac\alpha2\sin\frac\theta2\,\vec{v}'\cdot\vec{v}\right)+\\
&+\cos\frac\alpha2\sin\frac\theta2\vec{v}+\cos\frac\theta2\sin\frac\alpha2\vec{v}'+\sin\frac\alpha2\sin\frac\theta2\,\vec{v}'\times\vec{v}.
\end{aligned}
$$
From the second row we can read the axis of the composition - it is the unit vector parallel to that linear combination of $\vec{v}$, $\vec{v}'$ and their cross product. The first two terms are in the plane $T$ spanned by $\vec{v}$ and $\vec{v}'$, but the cross product is perpendicular to $T$. Therefore the axis of the combined rotation is in the plane $T$ if and only if that cross product term is zero. Either of the sines vanishes only when the rotation is trivial ($\alpha=0$ or $\theta=0$). The cross product vanishes iff $\vec{v}$ and $\vec{v}'$ are parallel.

In other words, your hunch is correct.
