(d) $ab = 6$
For parallel lines: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$
Here, $\dfrac{a}{3} = \dfrac{2}{b}$ $\Rightarrow$ $ab = 6$.
Source: Chapter 3, Section 3.2
For parallel lines, the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. Applying this to the given equations gives $\frac{a}{3} = \frac{2}{b}$, i.e., $ab = 6$. Note that option (d) is the only one expressing a product relation matching this cross-multiplication result.