1. **Problem Statement:** Given two planes with equations $\vec{r} \cdot \hat{a} = p$ and $\vec{r} \cdot \vec{b} - q = 0$, find: (a) The difference between the shortest distances of the planes from the origin $\vec{r} = \vec{0}$. 2. **Formula and Explanation:** The shortest distance $d$ from the origin to a plane $\vec{r} \cdot \vec{n} = d_0$ (where $\vec{n}$ is a unit normal vector) is given by the absolute value of the constant term divided by the magnitude of the normal vector. For plane 1: $\vec{r} \cdot \hat{a} = p$, since $\hat{a}$ is a unit vector, distance $d_1 = |p|$. For plane 2: $\vec{r} \cdot \vec{b} - q = 0$ can be rewritten as $\vec{r} \cdot \vec{b} = q$. The shortest distance is $d_2 = \frac{|q|}{|\vec{b}|}$. 3. **Intermediate Work:** Distance difference: $$\text{Difference} = |d_1 - d_2| = \left| |p| - \frac{|q|}{|\vec{b}|} \right|$$ 4. **Final Answer:** The difference between the shortest distances of the two planes from the origin is $$\boxed{\left| |p| - \frac{|q|}{|\vec{b}|} \right|}$$ --- Since the user message contains multiple problems, but per instructions, only the first problem is solved here.