The integer $$k$$, for which the inequality $$x^2 - 2(3k - 1)x + 8k^2 - 7 > 0$$ is valid for every $$x$$ in $$R$$ is:

For the quadratic expression $$x^2 - 2(3k - 1)x + 8k^2 - 7$$ to be strictly positive for every real $$x$$, the coefficient of $$x^2$$ must be positive (which it is, since it equals 1) and the discriminant must be strictly negative.

The discriminant is: $$D = [2(3k-1)]^2 - 4(1)(8k^2 - 7) = 4(3k-1)^2 - 4(8k^2 - 7)$$

Dividing by 4: $$\frac{D}{4} = (3k-1)^2 - (8k^2 - 7) = 9k^2 - 6k + 1 - 8k^2 + 7 = k^2 - 6k + 8$$

We need $$k^2 - 6k + 8 < 0$$. Factoring: $$(k - 2)(k - 4) < 0$$

This inequality holds when $$2 < k < 4$$. The only integer value of $$k$$ in this open interval is $$k = 3$$.

Therefore, the correct answer is Option 3: $$k = 3$$.