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

$\text{ If }(h,k)\text{ is foot of the perpendicular of }\left(x_1,y_1\right)\text{ on the line }ax+by+c=0\text{ then }$

$\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-\left(a{x}_1+b{y}_1+c\right)}{a^2+b^2}$

Hence, 

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

Therefore,

$\begin{array}{c}\frac{h+5}{3}=-1\\ h+5=-3\\ h=-8\end{array}$        And      $\begin{array}{c}\frac{k-7}{4}=-1\\ k-7=-4\\ k=3\end{array}$

Hence,

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

$\text{ Therefore, }\frac{h}{2k+2}=-1$