A particle travels in the X-Y plane governed by the equation r = (...

Given position vector $\vec{r} = (\hat{i} + 2\hat{j}) A \cos\omega t$

The $x$ and $y$ coordinates of the particle are:
$x = A \cos\omega t$
$y = 2A \cos\omega t$

Eliminating $t$, we get:
$y = 2x$
This is the equation of a straight line passing through the origin, so the particle moves on a straight line.

Differentiating $\vec{r}$ with respect to time $t$ to find acceleration:
$\vec{v} = \dfrac{d\vec{r}}{dt} = -(\hat{i} + 2\hat{j}) A \omega \sin\omega t$
$\vec{a} = \dfrac{d\vec{v}}{dt} = -(\hat{i} + 2\hat{j}) A \omega^2 \cos\omega t$
$\vec{a} = -\omega^2 \vec{r}$

Since the acceleration is directly proportional to the negative of the position vector, the motion is simple harmonic.

Every simple harmonic motion is inherently periodic.

Thus, the motion is periodic, on a straight line, and simple harmonic.