Let $$f(1) = -2$$ and $$f'(x) \geq 4.2$$ for $$1 \leq x \leq 6$$. The possible value of $$f(6)$$ lies in the interval :

We are given that $$ f(1) = -2 $$ and $$ f'(x) \geq 4.2 $$ for all $$ x $$ in the interval $$ [1, 6] $$. We need to find the interval in which the possible values of $$ f(6) $$ lie.

Since the function is differentiable on $$ [1, 6] $$ (as $$ f'(x) $$ is defined for all $$ x $$ in $$ [1, 6] $$), we can apply the mean value theorem. The mean value theorem states that there exists some point $$ c $$ in the open interval $$ (1, 6) $$ such that:

$$ f'(c) = \frac{f(6) - f(1)}{6 - 1} $$

Substituting the given values, $$ f(1) = -2 $$ and the interval length is $$ 6 - 1 = 5 $$, so:

$$ f'(c) = \frac{f(6) - (-2)}{5} = \frac{f(6) + 2}{5} $$

It is given that $$ f'(x) \geq 4.2 $$ for all $$ x $$ in $$ [1, 6] $$. Since $$ c $$ is in $$ (1, 6) $$, which is within $$ [1, 6] $$, we have $$ f'(c) \geq 4.2 $$. Therefore:

$$ \frac{f(6) + 2}{5} \geq 4.2 $$

To solve for $$ f(6) $$, multiply both sides of the inequality by 5:

$$ f(6) + 2 \geq 4.2 \times 5 $$

Calculating the right side:

$$ 4.2 \times 5 = 21 $$

So:

$$ f(6) + 2 \geq 21 $$

Subtract 2 from both sides:

$$ f(6) \geq 21 - 2 $$

$$ f(6) \geq 19 $$

This inequality means that $$ f(6) $$ is at least 19. Since the derivative $$ f'(x) $$ can be greater than 4.2 (as long as it is at least 4.2), $$ f(6) $$ can be larger than 19. There is no upper bound specified, so $$ f(6) $$ can be any number greater than or equal to 19. Therefore, the possible values of $$ f(6) $$ lie in the interval $$ [19, \infty) $$.

Now, comparing with the given options:

A. $$ [15, 19) $$

B. $$ (-\infty, 12) $$

C. $$ [12, 15) $$

D. $$ [19, \infty) $$

We see that option D matches the interval $$ [19, \infty) $$.

Hence, the correct answer is Option D.