Express the rotation of $\mathbb{R}^3$ by $\frac{\pi}{3}$ about the $x=y=z$ axis by using quaternions and identifying $\mathbb{R}^3$ with $(i,j,k)$-space.
Thoughts:
From my point of view, every point in $(i,j,k)$-space is a vector. Put the vector in $\mathbb{H}$ will get the real part of quaternion $=0$
Let $\theta$ be the angle of rotation in $(i,j,k)$-space about the axis $li+mj+nk$ coresponds to a quaternion part $\pm{q}$
$q=\cos{\frac{\theta}{2}} + (li+mj +nk) \sin{\frac{\theta}{2}}$
$\cos{\frac{\theta}{2}} = \cos{\frac{\frac{\pi}{3}}{2}} =\cos{\frac{\pi}{6}}=\frac{\sqrt3}{2}$
$\sin{\frac{\theta}{2}} = \sin{\frac{\frac{\pi}{3}}{2}} =\sin{\frac{\pi}{6}}=\frac{1}{2}$
$q=\frac{\sqrt3}{2} + \frac{1}{2} \hat{n}$
let $q\in\mathbb{H}$, we get pure quaternion $w=qvq^*$..
$|w|= |qvq^*|=|q||v||q^*|=|v|$....how should I continue???
In your calculation, $\hat n$ is the normalized and correctly oriented vector spanning the axis pf rotation. So in your case you have
\begin{align*} n &= \begin{pmatrix}1\\1\\1\end{pmatrix} & \hat n=\frac n{\lVert n\rVert}&=\frac1{\sqrt3}\begin{pmatrix}1\\1\\1\end{pmatrix} \end{align*}
and from that you get
$$q = \frac{\sqrt3}2 + \frac1{2\sqrt3}(i+j+k)$$.
You can verify that this is a rotation by checking
$$qq^* = \frac34 + \frac3{12} = \frac44 = 1$$
Then $v\mapsto w=qvq^*$ is the rotation described by $q$.