Let $|AB|=3$ and the length of the altitude from $C$ be $2$ in $\triangle ABC$. What is the maximum value of the product of the lengths of the other two altitudes?

$ 
\textbf{(A)}\ \frac{144}{25}
\qquad\textbf{(B)}\ 5
\qquad\textbf{(C)}\ 3\sqrt 2
\qquad\textbf{(D)}\ 6
\qquad\textbf{(E)}\ \text{None of the above}
$
1. Given that $|AB| = 3$ and the length of the altitude from $C$ is $2$ in $\triangle ABC$, we need to find the maximum value of the product of the lengths of the other two altitudes, $h_b$ and $h_c$.

2. Let $a = BC = 3$ and $h_a = 2$. The area $S$ of $\triangle ABC$ can be calculated using the formula for the area of a triangle:
   \[
   S = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 2 = 3
   \]

3. The lengths of the altitudes $h_b$ and $h_c$ can be expressed in terms of the area $S$ and the sides $b$ and $c$:
   \[
   h_b = \frac{2S}{b} \quad \text{and} \quad h_c = \frac{2S}{c}
   \]

4. The product of the altitudes $h_b$ and $h_c$ is:
   \[
   h_b h_c = \left(\frac{2S}{b}\right) \left(\frac{2S}{c}\right) = \frac{4S^2}{bc}
   \]

5. Using the area $S = 3$, we have:
   \[
   h_b h_c = \frac{4 \times 3^2}{bc} = \frac{36}{bc}
   \]

6. To maximize $h_b h_c$, we need to minimize $bc$. By the triangle inequality and the fact that $a = 3$, we know that $b + c > 3$ and $b, c > 0$.

7. For the maximum value of $h_b h_c$, consider the case when $\triangle ABC$ is isosceles with $AB = AC$. In this case, $b = c$.

8. Using the Pythagorean theorem in the isosceles triangle, we have:
   \[
   b^2 = \left(\frac{3}{2}\right)^2 + 2^2 = \frac{9}{4} + 4 = \frac{25}{4}
   \]
   \[
   b = \sqrt{\frac{25}{4}} = \frac{5}{2}
   \]

9. Therefore, $b = c = \frac{5}{2}$, and the product $bc$ is:
   \[
   bc = \left(\frac{5}{2}\right) \left(\frac{5}{2}\right) = \frac{25}{4}
   \]

10. Substituting $bc = \frac{25}{4}$ into the expression for $h_b h_c$:
    \[
    h_b h_c = \frac{36}{\frac{25}{4}} = 36 \times \frac{4}{25} = \frac{144}{25}
    \]

Conclusion:
\[
\boxed{\frac{144}{25}}
\]