(d) \(rac{23}{\sqrt{17}}\)The given lines are:L : \(rac{x}{5}+rac{y}{b}=1\) ....(i)K : \(rac{x}{c}+rac{y}{3}=1.\) ...(ii)Since line L passes through (13, 32),\(rac{13}{5}\) + \(rac{32}{b}\) = 1 ⇒ \(rac{32}{b}\) = 1 - \(rac{13}{5}\) = \(rac{-8}{5}\) ⇒ b = \(rac{32 imes5}{-8}\) = -20.∴ Line L is \(rac{x}{5}\) - \(rac{y}{20}\) = 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 = \(rac{-3x}{c}+3\)∴ Slope of K = \(-rac{3}{c}.\)given, \(-rac{3}{c}\) = 4 ⇒ c = \(-rac{3}{4}\)∴ Equation of line K : \(\big(rac{-4}{3}\big)x\) + \(rac{y}{3}\) = 1 ⇒ 4x – y + 3 = 0.Distance between the lines L ≡ 4x – y – 20 = 0 and K ≡ 4x – y + 3 = 0 isd = \(\bigg|rac{3-(-20)}{\sqrt{4^2+(-1)^2}}\bigg|\) = \(rac{23}{\sqrt{17}}\)\(\bigg(\)Distance between parallel lines ax + by + c1 = 0 and ax + by c2 = 0 is d = \(rac{|c_2-c_1|}{\sqrt{a^2+b^2}}\bigg)\)