The mean of the numbers $$a, b, 8, 5, 10$$ is 6 and the variance is 6.80. Then which one of the following gives possible values of $$a$$ and $$b$$?

Let the five numbers be $$a,\;b,\;8,\;5,\;10$$.

1. Using the definition of mean:
$$\text{Mean} = \frac{a+b+8+5+10}{5}=6$$

Simplifying,
$$a+b+23 = 30 \quad\Rightarrow\quad a+b = 7 \quad -(1)$$

2. For $$n$$ observations, the (population) variance is
$$\sigma^{2}= \frac{\sum_{i=1}^{n}(x_i-\bar{x})^{2}}{n}$$

Given variance $$= 6.80$$, mean $$=6$$ and $$n=5$$, we have
$$\frac{(a-6)^{2} + (b-6)^{2} + (8-6)^{2} + (5-6)^{2} + (10-6)^{2}}{5}=6.80$$

Calculate the squared deviations of the known numbers:
$$(8-6)^{2}=4,\qquad (5-6)^{2}=1,\qquad (10-6)^{2}=16$$
Sum of known terms $$=4+1+16=21$$

Let $$S=(a-6)^{2}+(b-6)^{2}$$. Then
$$\frac{S+21}{5}=6.80 \quad\Longrightarrow\quad S+21=34 \quad\Longrightarrow\quad S=13 \quad -(2)$$

3. From equations $$(1)$$ and $$(2)$$ we need
$$a+b=7,\qquad (a-6)^{2}+(b-6)^{2}=13$$

Check the given options:

Option A: $$a=0,\;b=7$$ gives $$a+b=7$$ but $$(0-6)^{2}+(7-6)^{2}=36+1=37\neq13$$

Option B: $$a=5,\;b=2$$ gives $$a+b=7$$ but $$(5-6)^{2}+(2-6)^{2}=1+16=17\neq13$$

Option C: $$a=1,\;b=6$$ gives $$a+b=7$$ but $$(1-6)^{2}+(6-6)^{2}=25+0=25\neq13$$

Option D: $$a=3,\;b=4$$ gives $$a+b=7$$ and $$(3-6)^{2}+(4-6)^{2}=9+4=13$$, satisfying both conditions.

Hence the only possible values are $$a=3,\;b=4$$.

Option D which is: $$a = 3,\ b = 4$$