The foot of the perpendicular from $\left(3,4\right)$  to $2x+3y+6=0$

$\frac{h-3}{2}=\frac{k-4}{3}=\frac{-\left(2\times 3+3\times 4+6\right)}{2^2+3^2}$

$\frac{h-3}{2}=\frac{k-4}{3}=\frac{-24}{13}$

$\frac{h-3}{2}=\frac{-24}{13}\Rightarrow h-3=\frac{-48}{13}\Rightarrow h=\frac{-9}{13}$

$\frac{k-4}{3}=\frac{-24}{13}\Rightarrow k-4=\frac{-72}{13}\Rightarrow k=\frac{-72}{13}+4=\frac{-20}{13}$

$\left(h,k\right)=\left(\frac{-9}{13},\frac{-20}{13}\right)$