The line $$L$$ given by $$\frac{x}{5} + \frac{y}{b} = 1$$ passes through the point $$(13, 32)$$. The line $$K$$ is parallel to $$L$$ and has the equation $$\frac{x}{c} + \frac{y}{3} = 1$$. Then the distance between $$L$$ and $$K$$ is

The line $$L$$ is given in intercept form as $$\frac{x}{5} + \frac{y}{b} = 1$$ and it passes through $$(13,\,32)$$.

Substituting the point to find $$b$$:
$$\frac{13}{5} + \frac{32}{b} = 1 \implies \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$$
$$\therefore\; b = \frac{32 \times 5}{-8} = -20$$

Hence the equation of $$L$$ is $$\frac{x}{5} - \frac{y}{20} = 1$$.
Multiplying by $$20$$ gives the general (ax + by + c = 0) form
$$4x - y - 20 = 0$$ $$-(1)$$

The parallel line $$K$$ is given as $$\frac{x}{c} + \frac{y}{3} = 1$$.
Convert to general form to compare slopes:
Multiply by $$3c$$: $$3x + cy = 3c$$
$$\Rightarrow y = -\frac{3}{c}x + 3$$

The slope of $$L$$ from $$-(1)$$ is $$4$$ (since $$y = 4x - 20$$).
For lines to be parallel, their slopes must be equal, so
$$-\frac{3}{c} = 4 \;\; \Longrightarrow \;\; c = -\frac{3}{4}$$

Substituting $$c = -\frac{3}{4}$$ back, the equation of $$K$$ becomes
$$-4x + y = 3 \;\; \text{or} \;\; 4x - y + 3 = 0$$ $$-(2)$$

The distance between two parallel lines $$ax + by + c_1 = 0$$ and $$ax + by + c_2 = 0$$ is
$$\displaystyle D = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}}$$

Here $$a = 4, \; b = -1, \; c_1 = -20 \;(\text{from } L), \; c_2 = 3 \;(\text{from } K).$$
$$D = \frac{|3 - (-20)|}{\sqrt{4^2 + (-1)^2}} = \frac{23}{\sqrt{16 + 1}} = \frac{23}{\sqrt{17}}$$

Therefore, the required distance is $$\frac{23}{\sqrt{17}}$$.
Option C which is: $$\frac{23}{\sqrt{17}}$$