$\text{ The lines }2x+5y=10\text{ and }3x+4y=12\text{ cuts the }x\text{-axis at }A\text{ and }B\text{ respectively. }A$

$\text{ line }L\text{ drawn through the point }(3,4)\text{ meets }y\text{-axis at }C\text{ such that the abscissae of }A$

$B\text{ and ordinate of }C\text{ are in geometric progression. If the equation of the line }L\text{ is }$

$ax+by+c=0\text{ then }a+b+c\text{ is }$

$\text{ The line }2x+5y=10\text{ cuts }x\text{-axis at }A(5,0)\text{ and the line }3x+4y=12\text{ cuts }x\text{-axis at }$$B\left(4,0\right)$

$\text{ The point Clies on }y\text{-axis, let it be }C(0,k)$

$\text{ Given }5,4,k\text{ are in geometric progression }$

It implies that 

$\frac{4}{5}=\frac{k}{4}$

Cross multiply 

$\begin{array}{c}5k=16\\ k=\frac{16}{5}\end{array}$

$\text{ Given }5,4,k\text{ are in geometric progression }$

$\text{ Equation of }L$

$\begin{array}{c}y-4=\frac{\frac{16}{5}-4}{0-3}\left(x-3\right)\\ =\frac{4}{15}\left(x-3\right)\\ 15y-60=4x-12\\ 4x-15y+48=0\end{array}$

$\text{ Hence, }l+m+n=4-15+48=37$