The number of integral values of $$k$$, for which one root of the equation $$2x^2 - 8x + k = 0$$ lies in the interval $$(1, 2)$$ and its other root lies in the interval $$(2, 3)$$, is :

We need to find integer values of $$k$$ such that one root of $$2x^2 - 8x + k = 0$$ lies in $$(1, 2)$$ and the other lies in $$(2, 3)$$.

Let $$f(x) = 2x^2 - 8x + k$$. For one root to lie in $$(1, 2)$$ and the other in $$(2, 3)$$, the function must change sign at the endpoints.

Evaluating $$f(x)$$ at the key points:

$$f(1) = 2(1)^2 - 8(1) + k = 2 - 8 + k = k - 6$$

$$f(2) = 2(2)^2 - 8(2) + k = 8 - 16 + k = k - 8$$

$$f(3) = 2(3)^2 - 8(3) + k = 18 - 24 + k = k - 6$$

For a root in $$(1, 2)$$: $$f(1)$$ and $$f(2)$$ must have opposite signs, so $$f(1) \cdot f(2) < 0$$.

$$(k - 6)(k - 8) < 0$$

This inequality holds when $$k$$ is between 6 and 8, i.e., $$6 < k < 8$$ $$-(1)$$

For a root in $$(2, 3)$$: $$f(2)$$ and $$f(3)$$ must have opposite signs, so $$f(2) \cdot f(3) < 0$$.

$$(k - 8)(k - 6) < 0$$

This also gives $$6 < k < 8$$ $$-(2)$$

From $$(1)$$ and $$(2)$$: $$k$$ must satisfy $$6 < k < 8$$.

The only integer in this open interval is $$k = 7$$.

Therefore, the number of integral values of $$k$$ is $$1$$, which is Option C.