What type of curve is it?

Let the acceleration be given by $X''=<1,1,1>\times X'.$ I will argue that this equation implies $X(t)$ is a circle. The acceleration is always perpendicular to $<1,1,1>.$ Therefore, the acceleration always lies in a plane (through the origin) perpendicular to the vector $<1,1,1>.$ If the initial velocity lies inside this same plane, then the velocity will always be confined to the plane. This would imply that the particle stays confined to some plane parallel to said plane. Note that the vector $<2,-1,-1>$ is perpendicular to $<1,1,1>$ , and, so, it lies inside said plane. Make this the initial velocity. Since $X''$ is also perpendicular to $X',$ we have $X'\bullet X''=0.$ This equation is equivalent to $|X'|^2=c,$ for some constant $c$ (verify by differentiating $X'\bullet X'=c$ and using the product rule for dot products). This means the curve has a constant speed. We also have $|X''|=|<1,1,1>||X'|\sin{\theta}=\sqrt{3}\sqrt{c}\sin\theta$ . I have already argued that the velocity is always perpendicular to $<1,1,1>.$ Therefore, $\sin\theta=1$ , and, so, the magnitude of the acceleration is also constant. These conditions describe a circle.

Now, note that $X''=<1,1,1>\times X'$ can be rewritten as

$X'' = \begin{bmatrix} 0 & -1 & 1\\ 1 & 0 & -1\\ -1 & 1 & 0 \end{bmatrix} X'$

The solution to this initial value problem (with $X(0)=<2,-1,-1>$ ) gives the result.