For the reaction $$2A + 3B + \frac{3}{2}C \to 3P$$, which statement is correct?

For any chemical reaction written in the stoichiometric form

$$aA + bB + cC \;\longrightarrow\; pP$$

the rate of reaction is defined by the relation

$$\text{Rate}\;=\;-\frac1a\,\frac{dn_A}{dt}\;=\;-\frac1b\,\frac{dn_B}{dt}\;=\;-\frac1c\,\frac{dn_C}{dt}\;=\;+\frac1p\,\frac{dn_P}{dt}$$

where $$dn_A/dt,\;dn_B/dt,\;dn_C/dt$$ are the time-rates of change of moles of the respective species. The negative signs for reactants indicate that their moles decrease with time, while the positive sign for products indicates an increase.

In the given reaction we have

$$2A + 3B + \frac32\,C \;\longrightarrow\; 3P$$

so the stoichiometric coefficients are $$a = 2,\; b = 3,\; c = \tfrac32,\; p = 3.$$ Substituting these values in the general definition, we get

$$\text{Rate}\;=\;-\frac12\,\frac{dn_A}{dt}\;=\;-\frac13\,\frac{dn_B}{dt}\;=\;-\frac2{3}\,\frac{dn_C}{dt}\;=\;+\frac13\,\frac{dn_P}{dt}.$$

Now we equate the first two terms to relate $$dn_A/dt$$ and $$dn_B/dt$$:

$$-\frac12\,\frac{dn_A}{dt}\;=\;-\frac13\,\frac{dn_B}{dt}.$$

Multiplying both sides by $$-1$$ removes the minus sign:

$$\frac12\,\frac{dn_A}{dt}\;=\;\frac13\,\frac{dn_B}{dt}.$$

Dividing both sides by $$\tfrac13$$ gives

$$\frac{dn_A}{dt}\;=\;\frac23\,\frac{dn_B}{dt}.$$

Next, equate the first and third expressions to connect $$dn_A/dt$$ and $$dn_C/dt$$:

$$-\frac12\,\frac{dn_A}{dt}\;=\;-\frac23\,\frac{dn_C}{dt}.$$

Again, cancelling the minus sign yields

$$\frac12\,\frac{dn_A}{dt}\;=\;\frac23\,\frac{dn_C}{dt}.$$

Multiplying both sides by $$2$$ gives

$$\frac{dn_A}{dt}\;=\;\frac43\,\frac{dn_C}{dt}.$$

Collecting the two results, we have

$$\frac{dn_A}{dt}\;=\;\frac23\,\frac{dn_B}{dt}\;=\;\frac43\,\frac{dn_C}{dt}.$$

This set of equalities is exactly what is written in Option C.

Hence, the correct answer is Option C.