Explanation

(d)
Since, line $L$ passes through (13, 32)
$\therefore\frac{13}{5} + \frac{32}{b} = 1$
$\Rightarrow \frac{32}{b} = 1 - \frac{13}{5} = - \frac{8}{5}$
$\Rightarrow b = - \frac{32 \times 5}{8} = - 20$
$\Rightarrow L:\frac{x}{5} - \frac{y}{20} = 1$
$Given,\ K:\frac{x}{c} + \frac{y}{3} = 1\ is\ parallel\ to\ L = 0$
$\therefore$ The line $K$ must have equation
$\frac{x}{5} - \frac{y}{20} = a$
$or\ \frac{x}{5a} - \frac{y}{20a} = 1$
$Comparing\ with\ \frac{x}{c} + \frac{y}{3} = 1$
$\Rightarrow \ - 20a = 3,\ c = 5a$
$\Rightarrow a = - \frac{3}{20},\ c = - \frac{15}{20}$
$\therefore\$Distance between line is
$\left| \frac{a - 1}{\sqrt{\frac{1}{25} + \frac{1}{400}}} \right| = \left| \frac{- \frac{3}{20} - 1}{\sqrt{\frac{17}{400}}} \right| = \frac{23}{\sqrt{17}}$