Let the mean and variance of the frequency distribution
$$x$$:       $$x_1 = 2$$       $$x_2 = 6$$       $$x_3 = 8$$        $$x_4 = 9$$
$$f$$:             4                  4                  $$\alpha$$                   $$\beta$$
be 6 and 6.8 respectively. If $$x_3$$ is changed from 8 to 7, then the mean for the new data will be:

First we write down the information given for the original frequency distribution.

The values are $$x_1 = 2,\;x_2 = 6,\;x_3 = 8,\;x_4 = 9$$ and the corresponding frequencies are $$f_1 = 4,\;f_2 = 4,\;f_3 = \alpha,\;f_4 = \beta.$$

We are told that the mean of this distribution is $$6.$$ By definition, the mean (denoted by $$\bar{x}$$) of a discrete frequency distribution is

$$\bar{x} \;=\; \dfrac{\displaystyle\sum_{i=1}^{4} f_i x_i}{\displaystyle\sum_{i=1}^{4} f_i}.$$

Substituting the known numbers we obtain

$$6 \;=\; \dfrac{4 \cdot 2 \;+\; 4 \cdot 6 \;+\; \alpha \cdot 8 \;+\; \beta \cdot 9}{4 + 4 + \alpha + \beta}.$$

Simplifying the numerator and denominator separately, we have

Numerator: $$4\cdot 2 = 8,\;\;4\cdot 6 = 24$$ so $$8 + 24 = 32.$$ Hence the numerator is $$32 + 8\alpha + 9\beta.$$

Denominator: $$4 + 4 = 8,$$ so the total frequency is $$8 + \alpha + \beta.$$

Therefore the mean equation becomes

$$6 \;=\; \dfrac{32 + 8\alpha + 9\beta}{8 + \alpha + \beta}.$$

Cross-multiplying gives

$$32 + 8\alpha + 9\beta \;=\; 6\,(8 + \alpha + \beta).$$

Expanding the right-hand side, $$6(8 + \alpha + \beta) = 48 + 6\alpha + 6\beta.$$

Now we bring all terms to one side:

$$32 + 8\alpha + 9\beta - 48 - 6\alpha - 6\beta = 0.$$

Combining like terms,

$$(8\alpha - 6\alpha) + (9\beta - 6\beta) + (32 - 48) = 0,$$ $$2\alpha + 3\beta - 16 = 0,$$ so

$$2\alpha + 3\beta = 16. \qquad (1)$$

Next we use the given variance. For a frequency distribution, the variance $$\sigma^{2}$$ is given by

$$\sigma^{2} \;=\; \dfrac{\displaystyle\sum_{i=1}^{4} f_i\,(x_i - \bar{x})^{2}}{\displaystyle\sum_{i=1}^{4} f_i}.$$

We are told that $$\sigma^{2} = 6.8.$$ Since the mean is $$\bar{x}=6,$$ we compute each squared deviation:

For $$x_1 = 2: (2-6)^2 = (-4)^2 = 16,$$ for $$x_2 = 6: (6-6)^2 = 0,$$ for $$x_3 = 8: (8-6)^2 = 2^2 = 4,$$ for $$x_4 = 9: (9-6)^2 = 3^2 = 9.$$

Multiplying by their frequencies gives

$$f_1(2-6)^2 = 4 \cdot 16 = 64,$$ $$f_2(6-6)^2 = 4 \cdot 0 = 0,$$ $$f_3(8-6)^2 = \alpha \cdot 4 = 4\alpha,$$ $$f_4(9-6)^2 = \beta \cdot 9 = 9\beta.$$

Hence the numerator of the variance expression is

$$64 + 4\alpha + 9\beta.$$

The denominator is the same total frequency $$8 + \alpha + \beta.$$

Thus the variance equation is

$$6.8 \;=\; \dfrac{64 + 4\alpha + 9\beta}{8 + \alpha + \beta}.$$

Cross-multiplying once again,

$$64 + 4\alpha + 9\beta \;=\; 6.8\,(8 + \alpha + \beta).$$

Compute $$6.8 \times 8 = 54.4,$$ so the right side is $$54.4 + 6.8\alpha + 6.8\beta.$$

Bringing all terms to the left gives

$$64 + 4\alpha + 9\beta - 54.4 - 6.8\alpha - 6.8\beta = 0.$$

Combining like terms,

$$(4\alpha - 6.8\alpha) + (9\beta - 6.8\beta) + (64 - 54.4) = 0,$$ $$-2.8\alpha + 2.2\beta + 9.6 = 0.$$

To clear the decimals, multiply every term by $$10$$:

$$-28\alpha + 22\beta + 96 = 0.$$

Dividing by $$2$$ for simplicity gives

$$-14\alpha + 11\beta + 48 = 0,$$ or equivalently

$$11\beta - 14\alpha = -48. \qquad (2)$$

Now we solve the simultaneous linear equations (1) and (2). From equation (1) we isolate $$\alpha$$:

$$2\alpha + 3\beta = 16 \;\;\Rightarrow\;\; 2\alpha = 16 - 3\beta \;\;\Rightarrow\;\; \alpha = 8 - \dfrac{3}{2}\beta.$$

Substituting this value of $$\alpha$$ into equation (2):

$$11\beta - 14\Bigl(8 - \dfrac{3}{2}\beta\Bigr) = -48.$$

First expand the term inside:

$$14\Bigl(8 - \dfrac{3}{2}\beta\Bigr) = 14 \cdot 8 - 14 \cdot \dfrac{3}{2}\beta = 112 - 21\beta.$$

Hence equation (2) becomes

$$11\beta - \bigl(112 - 21\beta\bigr) = -48.$$

Removing the parentheses carefully:

$$11\beta - 112 + 21\beta = -48.$$

Combine the $$\beta$$ terms:

$$(11\beta + 21\beta) - 112 = -48,$$ $$32\beta - 112 = -48.$$

Add $$112$$ to both sides:

$$32\beta = 64 \;\;\Rightarrow\;\; \beta = \dfrac{64}{32} = 2.$$

Now we substitute $$\beta = 2$$ back into $$\alpha = 8 - \dfrac{3}{2}\beta$$:

$$\alpha = 8 - \dfrac{3}{2}\times 2 = 8 - 3 = 5.$$

Thus we have determined the missing frequencies:

$$\alpha = 5,\quad \beta = 2.$$

Next, the problem states that the value $$x_3$$ is changed from $$8$$ to $$7.$$ Therefore the revised distribution is

$$x: 2,\;6,\;7,\;9$$ with the same frequencies $$4,\;4,\;5,\;2$$ respectively.

We now compute the mean of this new data set. The total frequency remains

$$N = 4 + 4 + 5 + 2 = 15.$$

The sum of the products $$f_i x_i$$ for the new data is

$$\begin{aligned} 4\cdot 2 &= 8,\\ 4\cdot 6 &= 24,\\ 5\cdot 7 &= 35,\\ 2\cdot 9 &= 18. \end{aligned}$$

Adding these contributions together,

$$8 + 24 = 32,\quad 32 + 35 = 67,\quad 67 + 18 = 85.$$

Hence the new mean $$\bar{x}_{\text{new}}$$ is

$$\bar{x}_{\text{new}} \;=\; \dfrac{85}{15} = \dfrac{17}{3}.$$

This value simplifies to $$5\dfrac{2}{3},$$ but writing it as an exact fraction clearly matches the choice given in the options.

Hence, the correct answer is Option C.