Take an orthonormal basis $(i,j,k)$ of $\mathbb R^3$, $v_1

Take an orthonormal basis $(i,j,k)$ of $\mathbb R^3$, $v_1 = \cos \frac{\pi}{6} i + \sin \frac{\pi}{6} j$ and $v_2 = \cos \frac{\pi}{6} i - \sin \frac{\pi}{6} j$. We have by construction $\angle(v_1, v_2) = \frac{\pi}{3}$.

Now let's find $\alpha$ such that $v_3 = \cos \alpha i + \sin \alpha k$ solve the problem.

That will be the case providing that $\angle(v_1,v_3) = \frac{\pi}{3}$, i.e. if $\cos \frac{\pi}{6} \cos \alpha = \cos \frac{\pi}{3}$, i.e. $\cos \alpha = \frac{1}{\sqrt 3}$ and $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \frac{\sqrt2}{\sqrt{3}}$.

Finally $$\begin{cases} v_1 &= \frac{\sqrt 3}{2} i + \frac{1}{2}j\\ v_2 &= \frac{\sqrt 3}{2} i - \frac{1}{2}j\\ v_3 &= \frac{1}{\sqrt 3}(i + \sqrt 2 k) \end{cases}$$

is a solution.