A particle travels along a straight line with a uniform velocity....

The angular momentum of a particle about a reference point is given by $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.

The magnitude of the angular momentum is $L = mvr\sin\theta = mvd$, where $d$ is the perpendicular distance from the reference point to the line of motion.

Since the particle moves with uniform velocity, its speed $v$ is constant and its line of motion is fixed. For any given fixed reference point, the perpendicular distance $d$ remains constant.

The direction of $\vec{L}$ is determined by the right-hand rule and also remains constant. Thus, the angular momentum vector remains constant about any given point.

If the reference point lies on the line of motion, $d = 0$, which gives $L = 0$.

If the reference point is away from the line of motion, $d \neq 0$, which gives $L \neq 0$.

Therefore, the angular momentum is non-zero about a point away from the straight line, zero about a point on the straight line, and remains constant about any given point.