We have,
$A(-5,-4)$ and slope of $L=0$ is $\tan \theta$
$\therefore \quad B=\left(-5+r_1 \cos \theta,-4+r_1 \sin \theta\right)$
and $\quad C=\left(-5+r_2 \cos \theta,-4+r_2 \sin \theta\right)$
Since , $B$ lies on $x+3 y+2=0$
$\therefore \quad-5+r_1 \cos \theta-12+3 r_1 \sin \theta+2=0$
$\begin{array}{ll}\Rightarrow & r_1=\frac{15}{\cos \theta+3 \sin \theta} \\ \therefore & A B=r_1=\frac{15}{\cos \theta+3 \sin \theta}\end{array}$
And $C$ lies on $2 x+y+4=0$
$\begin{aligned} & \therefore-10+2 r_2 \cos \theta-4+r_2 \sin \theta+4=0 \\ & \Rightarrow \quad r_2=\frac{10}{2 \cos \theta+\sin \theta} \\ & \therefore \quad\end{aligned}$
$\therefore \quad A C=r_2=\frac{10}{2 \cos \theta+\sin \theta}$
Now, we have
$\begin{gathered}\quad \frac{100}{A C^2}-\frac{225}{A B^2}=4 \cos 2 \theta+\sin 2 \theta \\ \Rightarrow \quad(2 \cos \theta+\sin \theta)^2-(\cos \theta+3 \sin \theta)^2 \\ =4 \cos 2 \theta+\sin 2 \theta \\ \Rightarrow \quad 4 \cos ^2 \theta+\sin ^2 \theta+4 \sin \theta \cos \theta-\cos ^2 \theta \\ \quad-9 \sin ^2 \theta-6 \sin \theta \cos \theta \\ \quad=4\left(\cos ^2 \theta-\sin ^2 \theta\right)+2 \sin \theta \cos \theta \\ \Rightarrow \quad 3 \cos ^2 \theta-8 \sin ^2 \theta-2 \sin \theta \cos \theta \\ \quad=4 \cos ^2 \theta-4 \sin { }^2 \theta+2 \sin \theta \cos \theta \\ \Rightarrow \quad 4 \sin ^2 \theta+\cos ^2 \theta+4 \sin \theta \cos \theta=0 \\ \Rightarrow \quad 4 \tan ^2 \theta+1+4 \tan \theta=0 \Rightarrow(2 \tan \theta+1)^2=0 \\ \Rightarrow \quad \tan \theta=-\frac{1}{2}\end{gathered}$
$\therefore$ Slope of line $L=\frac{-1}{2}$