I was trying to give a geometrical description of the the motion of Q, and the motion of Q can be described by the equation: $Q=cos(t+\frac{1}{4}\pi )[\frac{3}{2}\overrightarrow{i}+\frac{3\sqrt{3}}{2}\overrightarrow{k}]+3sin(t+\frac{1}{4}\pi )\overrightarrow{j}$

By calculating |Q|, we know that the distance from the origin is constant; hence, the particle would travel at a circular path.After proving that, I took a look at the mark scheme for this question, and the mark scheme indicated that:”it is evident that $\sqrt{3}x-z=0$, and so this defines the plane in which the motion of Q takes place”. I can’t understand what $\sqrt{3}x-z=0$ tells us, and why does it relate to the plane of motion, from my point of view, we can see that the equation of motion has variables in x-y-z, so it is obviously moves in x-y-z.

By calculating |Q|, we know that the distance from the origin is constant; hence, the particle would travel at a circular path.After proving that, I took a look at the mark scheme for this question, and the mark scheme indicated that:”it is evident that $\sqrt{3}x-z=0$, and so this defines the plane in which the motion of Q takes place”. I can’t understand what $\sqrt{3}x-z=0$ tells us, and why does it relate to the plane of motion, from my point of view, we can see that the equation of motion has variables in x-y-z, so it is obviously moves in x-y-z.