= Solution
The <straight line> consists of $\mathbf y+s(\mathbf z-\mathbf y)$, with $s\in\mathbb R$. Since the endpoints are distinct, the direction <vector> can be normalized. Choose
$$
\boxed{\mathbf p=\mathbf y,\qquad
\hat{\mathbf q}=\frac{\mathbf z-\mathbf y}{|\mathbf z-\mathbf y|}}.
$$
Putting $t=s|\mathbf z-\mathbf y|$ gives the required <parametric equation of a straight line> $\mathbf r=\mathbf p+t\hat{\mathbf q}$. Other choices of $\mathbf p$ on the same <straight line>, or the opposite direction for $\hat{\mathbf q}$, give equivalent descriptions.
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