The locus of a point equidistant from two intersecting lines is the pair of their angle bisectors.

Given lines: $L_1: x+2y+7=0,$ $L_2: 2x-y+8=0.$

The condition of equal distance is $\frac{|x+2y+7|}{\sqrt{1^2+2^2}}=\frac{|2x-y+8|}{\sqrt{2^2+(-1)^2}}.$

Since both denominators are equal to $\sqrt{5}$, we get $|x+2y+7|=|2x-y+8|.$

Squaring, $(x+2y+7)^2=(2x-y+8)^2.$

Use difference of squares: $[(x+2y+7)-(2x-y+8)]\,[(x+2y+7)+(2x-y+8)]=0.$

Simplify each factor: $x+2y+7-(2x-y+8)=-x+3y-1,$ $x+2y+7+2x-y+8=3x+y+15.$

So the locus is $(-x+3y-1)(3x+y+15)=0.$

Expand: \begin{align*} (-x+3y-1)(3x+y+15) &= -3x^2-xy-15x+9xy+3y^2+45y-3x-y-15 \ &= -3x^2+8xy+3y^2-18x+44y-15. \end{align*}

Multiplying by $-1$, $3x^2-8xy-3y^2+18x-44y+15=0.$

Compare with the general pair-of-lines form $ax^2+2hxy+by^2+2gx+2fy+c=0.$

From $3x^2-8xy-3y^2+18x-44y+15=0,$ we get $a=3,\quad 2h=-8\Rightarrow h=-4,\quad b=-3,$ $2g=18\Rightarrow g=9,\quad 2f=-44\Rightarrow f=-22,\quad c=15.$

Now compute: $g+c+h-f=9+15-4-(-22)=9+15-4+22=42.$

This does not match any option. So let us rewrite the equation in the exact form given in the question: $x^2-y^2+2hxy+2gx+2fy+c=0.$

Here the coefficients of $x^2$ and $y^2$ are fixed as $1$ and $-1$. So divide the obtained equation by $3$: $x^2-\frac{8}{3}xy-y^2+6x-\frac{44}{3}y+5=0.$

Hence, $2h=-\frac{8}{3}\Rightarrow h=-\frac{4}{3},$ $2g=6\Rightarrow g=3,$ $2f=-\frac{44}{3}\Rightarrow f=-\frac{22}{3},$ $c=5.$

Therefore, \begin{align*} g+c+h-f &=3+5-\frac{4}{3}-\left(-\frac{22}{3}\right)\ &=8+\frac{18}{3}\ &=14. \end{align*}

Therefore the correct option is $\boxed{14}.$