If the product of two zeros of the polynomial f(x) = 2x3 + 6x2 − 4x + 9 is 3,

Let $\alpha, \beta, \gamma$ be the zeros of polynomial $f(x)=2 x^{3}+6 x^{2}-4 x+9$ such that $\alpha \beta=3$

We have,

$\alpha \beta \gamma=\frac{\text { Constant term }}{\text { Coefficient of } x^{3}}$

$=\frac{-9}{2}$

Putting $\alpha \beta=3$ in $\alpha \beta \gamma=\frac{-9}{2}$, we get

$\alpha \beta \gamma=\frac{-9}{2}$

$3 \gamma=\frac{-9}{2}$

$y=\frac{-9}{2} \times \frac{1}{3}$

$\gamma=\frac{-3}{2}$

Therefore, the value of third zero is $\frac{-3}{2}$.

Hence, the correct alternative is $(b)$.