(d) \(\frac{23}{\sqrt{17}}\)

The given lines are:

L : \(\frac{x}{5}+\frac{y}{b}=1\)            ....(i)

K : \(\frac{x}{c}+\frac{y}{3}=1.\)          ...(ii)

Since line L passes through (13, 32),

\(\frac{13}{5}\) + \(\frac{32}{b}\) = 1 ⇒ \(\frac{32}{b}\) = 1 - \(\frac{13}{5}\) = \(\frac{-8}{5}\) ⇒ b = \(\frac{32\times5}{-8}\) = -20.

∴ Line L is \(\frac{x}{5}\) - \(\frac{y}{20}\) = 1, i.e., y = 4x – 20

⇒ Slope of line L = 4 

As L || K, slope of line K = Slope of line L. 

Eqn of line K can be written as y = \(\frac{-3x}{c}+3\)

∴ Slope of K = \(-\frac{3}{c}.\)

given, \(-\frac{3}{c}\) = 4 ⇒ c = \(-\frac{3}{4}\)

∴ Equation of line K : \(\big(\frac{-4}{3}\big)x\) + \(\frac{y}{3}\) = 1 ⇒ 4x – y + 3 = 0.

Distance between the lines 

L ≡ 4x – y – 20 = 0 and K ≡ 4x – y + 3 = 0 is

d = \(\bigg|\frac{3-(-20)}{\sqrt{4^2+(-1)^2}}\bigg|\) = \(\frac{23}{\sqrt{17}}\)
\(\bigg(\)Distance between parallel lines ax + by + c₁ = 0 and ax + by c₂ = 0 is d = \(\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}\bigg)\)