If a line L is perpendicular to the line $$5x - y = 1$$, and the area of the triangle formed by the line L and the coordinate axes is 5 sq units, then the distance of the line L from the line $$x + 5y = 0$$ is:

First, we need to find the slope of the given line $$5x - y = 1$$. To do this, we rearrange it into slope-intercept form $$y = mx + c$$.

Starting with $$5x - y = 1$$:

Subtract $$5x$$ from both sides: $$-y = -5x + 1$$

Multiply both sides by $$-1$$: $$y = 5x - 1$$

So, the slope of this line is 5.

Since line L is perpendicular to this line, the product of their slopes must be $$-1$$. Let the slope of L be $$m$$. Then:

$$m \times 5 = -1$$

Solving for $$m$$: $$m = -\frac{1}{5}$$

Thus, the slope of line L is $$-\frac{1}{5}$$.

The equation of line L can be written as $$y = -\frac{1}{5}x + b$$, where $$b$$ is the y-intercept. To avoid fractions, multiply through by 5:

$$5y = -x + 5b$$

Rearrange to standard form: $$x + 5y = 5b$$

Let $$c = 5b$$, so the equation becomes:

$$x + 5y = c$$     (1)

Now, we find the intercepts of line L with the coordinate axes to use the area condition.

When $$x = 0$$: $$5y = c$$ $$\Rightarrow$$ $$y = \frac{c}{5}$$   (y-intercept)

When $$y = 0$$: $$x = c$$   (x-intercept)

The area of the triangle formed with the axes is given as 5 sq units. The area formula is $$\frac{1}{2} \times |\text{x-intercept}| \times |\text{y-intercept}|$$. Since area is positive, we use absolute values:

$$\frac{1}{2} \times |c| \times \left|\frac{c}{5}\right| = 5$$

Since $$\left|\frac{c}{5}\right| = \frac{|c|}{5}$$, this becomes:

$$\frac{1}{2} \times |c| \times \frac{|c|}{5} = 5$$

Simplify: $$\frac{1}{2} \times \frac{|c|^2}{5} = 5$$

Because $$|c|^2 = c^2$$, we have:

$$\frac{c^2}{10} = 5$$

Multiply both sides by 10: $$c^2 = 50$$

Take square roots: $$c = \sqrt{50}$$ or $$c = -\sqrt{50}$$

Simplify: $$c = 5\sqrt{2}$$ or $$c = -5\sqrt{2}$$

So, the possible equations for line L are:

$$x + 5y = 5\sqrt{2}$$     or     $$x + 5y = -5\sqrt{2}$$

Now, we need to find the distance from line L to the line $$x + 5y = 0$$.

Write both lines in the form $$ax + by + k = 0$$:

The reference line is $$x + 5y = 0$$, so $$x + 5y + 0 = 0$$.

For the first case, $$x + 5y - 5\sqrt{2} = 0$$.

For the second case, $$x + 5y + 5\sqrt{2} = 0$$ (since $$x + 5y = -5\sqrt{2}$$ implies $$x + 5y + 5\sqrt{2} = 0$$).

The distance $$d$$ between two parallel lines $$ax + by + c_1 = 0$$ and $$ax + by + c_2 = 0$$ is:

$$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$$

Here, $$a = 1$$, $$b = 5$$, so $$\sqrt{a^2 + b^2} = \sqrt{1^2 + 5^2} = \sqrt{26}$$.

For the first line L: $$x + 5y - 5\sqrt{2} = 0$$, so $$c_1 = -5\sqrt{2}$$. The reference line has $$c_2 = 0$$.

Distance: $$d_1 = \frac{|-5\sqrt{2} - 0|}{\sqrt{26}} = \frac{5\sqrt{2}}{\sqrt{26}}$$

For the second line L: $$x + 5y + 5\sqrt{2} = 0$$, so $$c_1 = 5\sqrt{2}$$.

Distance: $$d_2 = \frac{|5\sqrt{2} - 0|}{\sqrt{26}} = \frac{5\sqrt{2}}{\sqrt{26}}$$

Both distances are the same. Simplify:

$$\frac{5\sqrt{2}}{\sqrt{26}} = 5 \sqrt{\frac{2}{26}} = 5 \sqrt{\frac{1}{13}} = \frac{5}{\sqrt{13}}$$

Alternatively, $$\frac{5\sqrt{2}}{\sqrt{26}} = \frac{5\sqrt{2}}{\sqrt{2 \times 13}} = \frac{5\sqrt{2}}{\sqrt{2} \times \sqrt{13}} = \frac{5}{\sqrt{13}}$$

So, the distance is $$\frac{5}{\sqrt{13}}$$ units.

Comparing with the options:

A. $$\frac{7}{\sqrt{13}}$$

B. $$\frac{7}{\sqrt{5}}$$

C. $$\frac{5}{\sqrt{13}}$$

D. $$\frac{5}{\sqrt{7}}$$

Hence, the correct answer is Option C.