Let $$L$$ denote the line in the $$xy$$-plane with $$x$$ and $$y$$ intercepts as 3 and 1 respectively. Then the image of the point $$(-1, -4)$$ in the line is:

We start by writing the equation of the given line. A line whose intercepts on the coordinate axes are $$a$$ on the $$x$$-axis and $$b$$ on the $$y$$-axis can be expressed in the intercept form $$\dfrac{x}{a} + \dfrac{y}{b} = 1$$. Here the intercepts are $$3$$ and $$1$$, so we have $$\dfrac{x}{3} + \dfrac{y}{1} = 1$$.

Multiplying every term by $$3$$ gives $$x + 3y = 3$$, which we rearrange to the standard form $$x + 3y - 3 = 0$$. Thus, for the line $$L$$ we identify $$a = 1$$, $$b = 3$$ and $$c = -3$$ in the general equation $$ax + by + c = 0$$.

To find the reflection (image) of a point $$(x_0,\,y_0)$$ across a line $$ax + by + c = 0$$, we use the well-known formula

$$ x' \;=\; x_0 \;-\; \dfrac{2a\,(ax_0 + by_0 + c)}{a^2 + b^2}, \qquad y' \;=\; y_0 \;-\; \dfrac{2b\,(ax_0 + by_0 + c)}{a^2 + b^2}. $$

For the point $$(-1,\,-4)$$ we substitute $$x_0 = -1$$ and $$y_0 = -4$$. First, we evaluate the expression $$ax_0 + by_0 + c$$:

$$ ax_0 + by_0 + c \;=\; (1)(-1) + (3)(-4) + (-3) \;=\; -1 - 12 - 3 \;=\; -16. $$

Next, we compute $$a^2 + b^2$$:

$$ a^2 + b^2 \;=\; 1^2 + 3^2 \;=\; 1 + 9 \;=\; 10. $$

Now we substitute everything into the reflection formulas.

For the reflected $$x$$-coordinate:

$$ x' \;=\; -1 \;-\; \dfrac{2 \cdot 1 \cdot (-16)}{10} \;=\; -1 \;+\; \dfrac{32}{10} \;=\; -1 + \dfrac{16}{5} \;=\; -\dfrac{5}{5} + \dfrac{16}{5} \;=\; \dfrac{11}{5}. $$

For the reflected $$y$$-coordinate:

$$ y' \;=\; -4 \;-\; \dfrac{2 \cdot 3 \cdot (-16)}{10} \;=\; -4 \;+\; \dfrac{96}{10} \;=\; -4 + \dfrac{48}{5} \;=\; -\dfrac{20}{5} + \dfrac{48}{5} \;=\; \dfrac{28}{5}. $$

Hence the image of $$(-1,\,-4)$$ in the line $$x + 3y - 3 = 0$$ is the point $$\left(\dfrac{11}{5},\,\dfrac{28}{5}\right)$$.

Among the given options, this corresponds to Option A.

Hence, the correct answer is Option A.