To solve this problem, let's break it down step by step :

Step 1 : Determine the mean of set C

The mean of set A = $5$

The mean of set B = $8$

After subtracting 3 from each element in set A, the new mean becomes $5 - 3 = 2$. After adding 2 to each element in set B, the new mean becomes $8 + 2 = 10$.

When we combine both modified sets to form set C, the mean of set C is the weighted average of the means of these modified sets:

Mean of C = $\frac{5 \times 2 + 5 \times 10}{10} = \frac{10 + 50}{10} = 6$

Step 2 : Determine the variance of set C

Variance is defined as the expectation of the squared deviation of a random variable from its mean. One property of variance is that, if you add (or subtract) a constant from each data point in a set, the variance of the set does not change.

Thus, the variance of the elements in set A remains $12$ even after subtracting 3 from each element, and the variance of the elements in set B remains $20$ even after adding 2 to each element.

Now, when combining variances from two datasets into one :

Variance of C

= $\frac{n_1 \times \text{Variance of A} + n_1 \times (\text{Mean of modified A} - \text{Mean of C})^2 + n_2 \times \text{Variance of B} + n_2 \times (\text{Mean of modified B} - \text{Mean of C})^2}{n_1 + n_2}$

Given :

$n_1 = n_2 = 5$

Variance of A = $12$, Variance of B = $20$

Mean of modified A = $2$, Mean of modified B = $10$, Mean of C = $6$

Plugging in the values, we get :

Variance of C = $\frac{5 \times 12 + 5 \times (2 - 6)^2 + 5 \times 20 + 5 \times (10 - 6)^2}{10}$

Variance of C = $\frac{60 + 80 + 100 + 80}{10} = \frac{320}{10} = 32$

Step 3 : Sum of the mean and variance of set C

Sum = Mean of C + Variance of C = $6 + 32 = 38$

So, the correct answer is :

Option C : 38.