The point where the perpendicular from the point $\left(-5,7\right)$ meets the line $3\mathrm{x}+4\mathrm{y}+12=0$ is $\left(\mathrm{h},\mathrm{k}\right)$ then the value of $\frac{\mathrm{h}}{2\mathrm{k}+2}$ is

If $\left(\mathrm{h},\mathrm{k}\right)$ is foot of the perpendicular of $\left(x_1,y_1\right)$ on the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ then  $\frac{\mathrm{h}-{\mathrm{x}}_1}{\mathrm{a}}=\frac{\mathrm{k}-{\mathrm{y}}_1}{\mathrm{b}}=\frac{-\left({\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}\right)}{{\mathrm{a}}^2+{\mathrm{b}}^2}$

Hence, 

 $\begin{array}{l}\frac{\mathrm{h}+5}{3}=\frac{\mathrm{k}-7}{4}=\frac{-\left(3\left(-5\right)+4\left(7\right)+12\right)}{3^2+4^2}\\ \frac{\mathrm{h}+5}{3}=\frac{\mathrm{k}-7}{4}=-\frac{25}{25}\\ \frac{\mathrm{h}+5}{3}=\frac{\mathrm{k}-7}{4}=-1\end{array}$           

Therefore,

$\begin{array}{c}\frac{\mathrm{h}+5}{3}=1\\ \mathrm{h}+5=3\\ \mathrm{h}=-2\end{array}$

 And    

$\begin{array}{c}\frac{\mathrm{k}-7}{4}=-1\\ \mathrm{k}-7=-4\\ \mathrm{k}=3\end{array}$

Hence,

$\begin{array}{c}\frac{\mathrm{h}}{2\mathrm{k}+2}=\frac{-2}{2\left(3\right)+2}\\ =-\frac{2}{8}\end{array}$

Therefore,$\frac{\mathrm{h}}{2\mathrm{k}+2}=\boxed{-0.25}$