Sure! A symmetric matrix is a square matrix that is equal to its transpose. For a matrix to be symmetric, the element at row i and column j must be equal to the element at row j and column i. In other words, $$A_{ij} = A_{ji}$$.

For a 3 $$ \times $$ 3 symmetric matrix, it looks like this:

$$ \begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \\ \end{pmatrix} $$

Notice that there are only 6 unique elements we need to fill because of the symmetry:

1. $$a$$ in the (1,1) position
2. $$b$$ in the (1,2) and (2,1) positions
3. $$c$$ in the (1,3) and (3,1) positions
4. $$d$$ in the (2,2) position
5. $$e$$ in the (2,3) and (3,2) positions
6. $$f$$ in the (3,3) position

Each of these unique elements can take a value from the set $${0,1,2,3,4,5,6,7,8,9}$$, which has 10 elements.

We have 10 choices for each of the 6 unique elements, so the total number of symmetric matrices can be calculated as:

$$10 \times 10 \times 10 \times 10 \times 10 \times 10 = 10^{6}$$

Thus, the total number of symmetric matrices of order 3 with entries from this set is $$10^{6}$$.