First, consider all three crates as a single system with total mass $M = m_1 + m_2 + m_3$. The net horizontal force on the system is the applied force $F$ minus the total kinetic friction force $f_k$.

Applying Newton's second law to the system:

$$F - f_k = Ma$$

The total friction force is $f_k = \mu_k N = \mu_k M g$.

$$F - \mu_k Mg = Ma$$

Solving for the system's acceleration $a$:

$$a = \frac{F - \mu_k Mg}{M} = \frac{F}{M} - \mu_k g$$

Next, apply Newton's second law to crate 3. The net force is the contact force from crate 2, $F_{32}$, minus the friction on crate 3, $f_{k,3}$.

$$F_{32} - f_{k,3} = m_3 a$$

The friction force on crate 3 is $f_{k,3} = \mu_k m_3 g$.

$$F_{32} = m_3 a + \mu_k m_3 g$$

Substitute the expression for acceleration $a$:

$$F_{32} = m_3 \left( \frac{F}{M} - \mu_k g \right) + \mu_k m_3 g$$ $$F_{32} = \frac{m_3 F}{M} - m_3 \mu_k g + \mu_k m_3 g$$ $$F_{32} = \frac{m_3 F}{M}$$

Substituting $M = m_1 + m_2 + m_3$, we find that $F_{32}$ is independent of the coefficient of friction $\mu_k$. Therefore, changing $\mu_k$ does not affect $F_{32}$.