For the reaction, 3A + 2B → C + D, the differential rate law can be written as:

For the reaction $$3A + 2B \rightarrow C + D$$, we need to write the differential rate law. The rate of a reaction is defined as the rate of disappearance of reactants or the rate of appearance of products, divided by their respective stoichiometric coefficients. This ensures the rate is consistent regardless of which species we measure.

The stoichiometric coefficients are 3 for A, 2 for B, 1 for C, and 1 for D. Therefore, the rate of the reaction (denoted as rate) can be expressed in multiple ways:

Rate $$ = -\frac{1}{3} \frac{d[A]}{dt} $$ because A is a reactant and its concentration decreases over time.

Rate $$ = -\frac{1}{2} \frac{d[B]}{dt} $$ because B is a reactant and its concentration decreases.

Rate $$ = +\frac{1}{1} \frac{d[C]}{dt} = \frac{d[C]}{dt} $$ because C is a product and its concentration increases.

Rate $$ = +\frac{1}{1} \frac{d[D]}{dt} = \frac{d[D]}{dt} $$ because D is a product and its concentration increases.

All these expressions represent the same rate. The differential rate law relates this rate to the concentrations of the reactants, typically written as $$ k [A]^n [B]^m $$, where $$ k $$ is the rate constant, and $$ n $$ and $$ m $$ are the orders of the reaction with respect to A and B.

We are asked to compare the expressions involving $$ \frac{d[A]}{dt} $$ and $$ \frac{d[C]}{dt} $$. From above:

Rate $$ = -\frac{1}{3} \frac{d[A]}{dt} $$

and

Rate $$ = \frac{d[C]}{dt} $$

Since both equal the rate, we set them equal:

$$ -\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} $$

Additionally, the rate is given by $$ k [A]^n [B]^m $$, so:

$$ -\frac{1}{3} \frac{d[A]}{dt} = k [A]^n [B]^m $$

and

$$ \frac{d[C]}{dt} = k [A]^n [B]^m $$

Therefore, we can write:

$$ -\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k [A]^n [B]^m $$

Now, let's examine the options:

Option A: $$ +\frac{1}{3} \frac{d[A]}{dt} = -\frac{d[C]}{dt} = k [A]^n [B]^m $$

This implies $$ +\frac{1}{3} \frac{d[A]}{dt} = k [A]^n [B]^m $$ and $$ -\frac{d[C]}{dt} = k [A]^n [B]^m $$. However, since A is a reactant, $$ \frac{d[A]}{dt} $$ is negative, so $$ +\frac{1}{3} \frac{d[A]}{dt} $$ would be negative, but $$ -\frac{d[C]}{dt} $$ would be negative (as $$ \frac{d[C]}{dt} $$ is positive), while the rate $$ k [A]^n [B]^m $$ is positive. This does not match our derived equation.

Option B: $$ \frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k [A]^n [B]^m $$

This implies $$ \frac{1}{3} \frac{d[A]}{dt} = k [A]^n [B]^m $$ and $$ \frac{d[C]}{dt} = k [A]^n [B]^m $$. Since A is decreasing, $$ \frac{d[A]}{dt} $$ is negative, so $$ \frac{1}{3} \frac{d[A]}{dt} $$ is negative, but $$ \frac{d[C]}{dt} $$ is positive. A negative quantity cannot equal a positive quantity unless both are zero, which is not generally true. Hence, this is incorrect.

Option C: $$ -\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k [A]^n [B]^m $$

This matches exactly what we derived: $$ -\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k [A]^n [B]^m $$.

Option D: $$ -\frac{d[A]}{dt} = \frac{d[C]}{dt} = k [A]^n [B]^m $$

This implies $$ -\frac{d[A]}{dt} = k [A]^n [B]^m $$ and $$ \frac{d[C]}{dt} = k [A]^n [B]^m $$. From our rate definition, $$ -\frac{d[A]}{dt} = 3 \times \left( -\frac{1}{3} \frac{d[A]}{dt} \right) = 3 \times \text{rate} $$, and $$ \frac{d[C]}{dt} = \text{rate} $$. So $$ -\frac{d[A]}{dt} = 3 \times \frac{d[C]}{dt} $$, meaning $$ -\frac{d[A]}{dt} $$ is three times $$ \frac{d[C]}{dt} $$. The option claims they are equal, which would require $$ 3 \times \frac{d[C]}{dt} = \frac{d[C]}{dt} $$, implying $$ \frac{d[C]}{dt} = 0 $$, which is not true in general. Hence, this is incorrect.

Therefore, the correct option is C.

Hence, the correct answer is Option C.