Amazing coordinate geometry

Let the required line $L$ make $\theta$ angle with $x-axis$ . See NOTE first.

Then distance of $B$ from $A(-5,-4)$ along $L$ :-

$\dfrac{|1(-5)+3(-4)+2|}{\cos\theta+3\sin \theta}=\color{#D61F06}{\dfrac{15}{\cos \theta +3\sin \theta}}$

Similarly distance of $C$ from $A(-5,-4)$ along $L$ :-

$\dfrac{|2(-5)+1(-4)+4|}{2\cos\theta+\sin \theta}=\color{#D61F06}{\dfrac{10}{2\cos \theta +\sin \theta}}$

And distance of $D$ from $A(-5,-4)$ along $L$ :-

$\dfrac{|1(-5)+(-1)(-4)+-5|}{\cos\theta+(-1)\sin \theta}=\color{#D61F06}{\dfrac{6}{\cos \theta-\sin \theta}}$

These are the values of $AB,AC,AD$ respectively which on direct substitution in the given equation gives:-

$\left(\dfrac{\cancel{15}}{\frac{\cancel{15}}{( \cos \theta +3\sin \theta)}}\right)^2+\left(\dfrac{\cancel{10}}{\frac{\cancel{10}}{ (2\cos \theta +\sin \theta)}}\right)^2= \left(\dfrac{\cancel 6}{\frac{\cancel 6}{(\cos \theta-\sin \theta)}}\right)^2$

$\implies 9\sin^2\theta +12\sin \theta\cos \theta+4\cos^2\theta=0$

$\implies 9\tan^2\theta+12\tan \theta+4=0$

$\implies (3\tan \theta+2)^2=0\implies \tan \theta=\dfrac{-2}3$

Hence equation of $L$ :

$(y+4)=\dfrac{-2}{3}(x+5)$

$\implies 2x+3y+22=0$

$\therefore 2+3+22=\boxed{27}$

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NOTE:- The distance of a point $(x_1,y_1)$ from a line $Ax+By+C=0$ along a particular direction(i.e along a line with slope $\tan \theta$ ) is given by : $\large \boxed{\left|\dfrac{Ax_1+By_1+C}{A\cos \theta+B\sin \theta}\right|}$