If the equation of the straight line passing through the point of intersection of $x+2y-19=0,x-2y-3=0$ and which is at a perpendicular distance of 5 units from the point (-2, 4) is $5x+by+c=0,$ then 5+b+c= ………..

Clearly point of intersection of the lines $x+2y-19=0,x-2y-3=0$ is P = (11, 4)

Let m be the slope of the line through P is $(y-4)=m(x-11)\Rightarrow mx-y+(4-11)m=0$

Distance from (-2, 4) to the above line is $5\Rightarrow \frac{|-2m-4+4-11m|}{\sqrt{m^2+1}}=5$

$\Rightarrow |-13m|=5\sqrt{m^2+1}\Rightarrow 169m^2=25(m^2+1)\Rightarrow 144m^2=25\Rightarrow m^2=\frac{25}{144}\Rightarrow m=\pm \frac{5}{12}$

If $m=\frac{5}{12},$ then the equation of the lines is $(y-4)=\frac{5}{12}(x-11)\Rightarrow 12y-48=5x-55$

$\Rightarrow 5x-12y-7=0.$ But it is given as $5x+by+c=0 \therefore b=-12,c=-7$

Now $5+b+c=5-12-7=5-19=-14$