$\text{ The equation of line belonging to the family of lines }p(2x+3y-13)+q(x-y+1)=0\text{ , }$

$\text{ and which is at maximum distance from the origin is }$

$\begin{array}{l}\text{ Given family of lines }p(2x+3y-13)+q(x-y+1)=0\\ \Rightarrow 2x+3y-13+\frac{q}{p}(x-y+1)=0\\ \Rightarrow 2x+3y-13+\lambda (x-y+1)=0\left(\because \text{ Let }\frac{p}{q}=\lambda \right)\\ \Rightarrow x(2+\lambda )+y(3-\lambda )+\lambda -13=0\dots \dots \dots ...\left(1\right)\\ \text{ The distance from origin to (1) is }D=\frac{|\lambda -13|}{\sqrt{(2+\lambda {)}^2+(3-\lambda {)}^2}}\end{array}$

$\begin{array}{l}=\frac{|\lambda -13|}{\sqrt{2{\lambda }^2-2\lambda +13}}\\ {\text{ Differentiate with respect to }}^{\mathrm{\prime }}{\lambda }^{\mathrm{\prime }}\\ \Rightarrow \frac{\mathrm{d}D}{\mathrm{d}\lambda }=\frac{\sqrt{2{\lambda }^2-2\lambda +13}-(\lambda -13)\frac{(4\lambda -2)}{2\left(\sqrt{2{\lambda }^2-2\lambda +13}\right)}}{\left(2{\lambda }^2-2\lambda +13\right)}\\ =\frac{2{\lambda }^2-2\lambda +13-(\lambda -13)(2\lambda -1)}{{\left(2{\lambda }^2-2\lambda +13\right)}^{\frac{3}{2}}}\end{array}$

For maximum value of D ,

$\begin{array}{l}\frac{\mathrm{d}D}{\mathrm{d}\lambda }=0\\ \Rightarrow 2{\lambda }^2-2\lambda +13=(\lambda -13)(2\lambda -1)\\ \Rightarrow \lambda =0\\ \text{ Substituting }\lambda =0\text{ in equation (1) we get }2x+3y-13=0\end{array}$