Step 1: Triple product
Compute $\vec{q}\times\vec{r}$ where $\vec{q} = (0, a, -a)$ and $\vec{r} = (3, a, 0)$. The components are $(a\cdot0 - (-a)\cdot a,\ (-a)\cdot3 - 0\cdot0,\ 0\cdot a - a\cdot3) = (a^2, -3a, -3a)$.
Step 2: Dot with p
$\vec{p} = (2a, 0, 1)$, so the product is $2a^3 + 0 - 3a$.
Step 3: Optimise
The magnitude $|2a^3 - 3a| = 3a - 2a^3$ on $[0, 1]$, which is largest where its derivative $3 - 6a^2 = 0$, so $a^2 = \frac{1}{2}$.
Step 4: Value
$3a - 2a^3 = a(3 - 2a^2) = \frac{1}{\sqrt{2}}\cdot2 = \sqrt{2}$.
Final Answer:
The maximum volume is sqrt 2. This is option (B).
\[ \boxed{\text{(B) }\sqrt{2}} \]