I had a test tonight and there was one problem that I couldn't even think about a solution. Problem was this, as much I can remember:
You are doing two rotations. First one from $y$-axis to $z$-axis and then from $x$-axis to $z$-axis. Question was: in what axis you can do rotation to replace this two rotations into only one rotation?
Answer as I can remember was the axis through the origin and $(1,-1,1)$. Would be very nice if you can explain in graphical way.
If the $y$ axis is rotated to the $z$ axis, the $x$ axis remains unchanged and the $z$ axis is rotated to the $-y$ axis. The matrix transformation for this is $$\left(\begin{matrix}1&0&0\\ 0&0&-1\\0&1&0\end{matrix}\right)$$
Then if the $x$ axis is rotated to the $z$ axis, the $y$ axis remains unchanged and the $z$ axis is rotated to the $-x$ axis. The matrix transformation for this is $$\left(\begin{matrix}0&0&-1\\0&1&0\\1&0&0\end{matrix}\right)$$
Multiplying these in the correct order gives the matrix $$\left(\begin{matrix}0&-1&0\\0&0&-1\\1&0&0\end{matrix}\right)$$
So $x$ goes to $z$, $y$ goes to $-x$ and $z$ goes to $-y$.
The rather poorly drawn diagram below shows the final result, with the axis of rotation in blue
EDIT: in response to the request for clarification from the OP, see the diagram below, where $-y$ goes to $x$ instead of $y$ goes to $-x$ (which is obviously the same). You can now see a complete circle in yellow which is in a plane perpendicular to the axis of rotation in blue. There is only one such possible axis.
Moreover it is easily verified, by calculating eigenvalues and eigenvectors that the axis of rotation is what you suspected, i.e. $$\left(\begin{matrix}1\\-1\\1\end{matrix}\right)$$