The point $(x,y)$ lies on the line $2x+3y=6$. The smallest value of the quantity $\sqrt{x^2+y^2}$ is m.  Then the value of $\sqrt{13}m$ is____

Let $x=r\cos \theta ,y=r\sin \theta ,r=\sqrt{x^2+y^2}$

$2r\cos \theta +3r\sin \theta =6\Rightarrow r=\frac{6}{2\cos \theta +3\sin \theta }$

$\because ris\min :$when $2\cos \theta +3\sin \theta$ must be max 

$\begin{array}{l}\therefore r_{\min }=\frac{6}{\sqrt{13}}=m\\ \sqrt{13}m=6\end{array}$