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 $\frac{x}{5}+\frac{y}{b}=1$ If passes through (13,32)

$\begin{array}{l}\Rightarrow \frac{13}{5}+\frac{32}{b}=1\\ \Rightarrow \frac{32}{b}=1-\frac{13}{5}\Rightarrow \frac{32}{b}=\frac{-8}{5}\\ \Rightarrow b=32\times \frac{-5}{8}\\ \Rightarrow b=-20\end{array}$

$\therefore Lis\frac{x}{5}+\frac{y}{-20}=1\Rightarrow 4x-y-20=0$

Slope of L= 4

               The line K is $\frac{x}{c}+\frac{y}{3}=1$

               Slope of $K=\frac{-3}{c}$

               $\therefore L{\parallel }^{le}K\Rightarrow$ slope of  L= slope of K

$\begin{array}{l}\Rightarrow 4=\frac{-3}{c}\Rightarrow c=\frac{-3}{4}\\ \therefore Kis\frac{x}{\left(\frac{-3}{4}\right)}+\frac{y}{3}=1\\ \Rightarrow 4x-y+3=0\end{array}$

Distance b/w L and K $=\frac{\left|-20-3\right|}{\sqrt{16+1}}=\frac{23}{\sqrt{17}}$