A line through  $A(-5,-4)$ with slope $\tan \theta$  meets the lines $x+3y+2=0$, $2x+y+4=0$, $x-y-5=0$ at B, C, D respectively, such that  
${\left(\frac{{\displaystyle 15}}{{\displaystyle AB}}\right)}^2+{\left(\frac{{\displaystyle 10}}{{\displaystyle AC}}\right)}^2={\left(\frac{{\displaystyle 6}}{{\displaystyle AD}}\right)}^2$then

A line through  $A(-5,-4)$ with slope $\tan \theta$ is $\frac{x+5}{\cos \theta }=\frac{y+4}{\sin \theta }=r$ 
Any point on the line is $=(-5+r\cos \theta ,-4+r\sin \theta )$
If this lies on $x+3y+2=0,$ we have $-5+r\cos \theta +3(-4+r\sin \theta )+2=0$

$\therefore r=\frac{15}{AB}=\cos \theta +3\sin \theta$
similarly, we get, $\frac{10}{AC}=2\cos \theta +\sin \theta$ and $\frac{6}{AD}=\cos \theta -\sin \theta$
From conditions, ${(\cos \theta +3\sin \theta )}^2+{(2\cos \theta +\sin \theta )}^2={(\cos \theta -\sin \theta )}^2$

$\Rightarrow {(2\cos \theta +3\sin \theta )}^2=0$

$\Rightarrow \tan \theta =-\frac{2}{3}$