The foot of the perpendicular on the line $3x+y=\lambda$  drawn from origin is  C, if the line meets the coordinate axes at $A\text{\hspace{0.17em}and\hspace{0.17em}}B$ are respectively, then $BC:CA=p:q$, then $p-q=$ _____

$\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-(a{x}_1+b{y}_1+c)}{a^2+b^2}$

Foot of perpendicular from origin to $3x+y=\lambda$

$\begin{array}{l}\frac{h-0}{3}=\frac{k-0}{1}=\frac{-\left[3\left(0\right)+0-\lambda \right]}{3^2+1^2}\\ \frac{h}{3}=\frac{k}{1}=\frac{\lambda }{10}\\ C=(h,k)=\left(\frac{3\lambda }{10},\frac{\lambda }{10}\right)\end{array}$

A is point on $X-$axis is $(x,0)=\left(\frac{\lambda }{3},0\right)$

B is point on $Y-$axis is $(0,y)=(0,\lambda )$

$\begin{array}{l}BC=\sqrt{{\left(\frac{3\lambda }{10}\right)}^2+{\left(\lambda -\frac{\lambda }{10}\right)}^2}\\ =\sqrt{\frac{9{\lambda }^2}{100}+\frac{81{\lambda }^2}{100}}=\sqrt{\frac{90{\lambda }^2}{100}}=\frac{3\lambda }{10}\sqrt{10}=\frac{3\lambda }{\sqrt{10}}\\ CA=\sqrt{{\left(\frac{\lambda }{3}-\frac{3\lambda }{10}\right)}^2+{\left(\frac{\lambda }{10}\right)}^2}\\ =\sqrt{{\left(\frac{10\lambda -9\lambda }{30}\right)}^2+\frac{{\lambda }^2}{100}}\\ =\sqrt{\frac{{\lambda }^2}{900}+\frac{{\lambda }^2}{100}}=\sqrt{\frac{{\lambda }^2+9{\lambda }^2}{900}}=\sqrt{\frac{10{\lambda }^2}{900}}\\ =\frac{\lambda }{3\sqrt{10}}\end{array}$

$BC:CA=\frac{3\lambda }{\sqrt{10}}:\frac{\lambda }{3\sqrt{10}}=9:1$

Here $p=9,q=1\Rightarrow p-q=8$