Set coordinates of the moving endpoints

Let $A=(a, frac{}{ }a+2)$ since $A$ lies on $x-y+2=0 \Rightarrow y=x+2$.

Let $B=(b,-2)$ since $B$ lies on $y+2=0 \Rightarrow y=-2$.

Use the rod length condition

The rod has fixed length $8$, so $AB^2=64.$

Now, $AB^2=(a-b)^2\,+\,(a+2-(-2))^2=(a-b)^2+(a+4)^2.$

Hence, $(a-b)^2+(a+4)^2=64.$

Coordinates of point $P$ dividing $AB$ internally in ratio $2:1$

Since $P$ divides $AB$ internally in the ratio $2:1$, we take $AP:PB=2:1.$

By section formula, $P=\left(\frac{2b+a}{3},\frac{2(-2)+(a+2)}{3}\right)=\left(\frac{a+2b}{3},\frac{a-2}{3}\right).$

Let $P=(x,y)$. Then $x=\frac{a+2b}{3},\qquad y=\frac{a-2}{3}.$

Express $a,b$ in terms of $x,y$

From $y=\frac{a-2}{3} \Rightarrow a=3y+2.$

Also, $x=\frac{a+2b}{3} \Rightarrow 3x=a+2b \Rightarrow 2b=3x-a=3x-(3y+2),$ so $b=\frac{3x-3y-2}{2}.$

Substitute into the length equation

First, a-b=(3y+2)-\frac{3x-3y-2}{2}=rac{9y+6-3x}{2}.

Also, $a+4=3y+6=3(y+2).$

Therefore, $\left(\frac{9y+6-3x}{2}\right)^2+[3(y+2)]^2=64.$

Simplify: $\frac{9}{4}(3y+2-x)^2+9(y+2)^2=64.$

Multiply by $4$: $9(3y+2-x)^2+36(y+2)^2=256.$

Divide by $4$ alternatively from earlier is easier: $\frac{9}{4}(3y+2-x)^2+9(y+2)^2=64.$ Multiply by $4$: $9(3y+2-x)^2+36(y+2)^2=256.$

Divide by $4$? No need. Expand directly:

$9(x-3y-2)^2+36(y+2)^2=256.$

Expanding, $9(x^2+9y^2+4-6xy-4x+12y)+36(y^2+4y+4)=256.$

$9x^2+81y^2+36-54xy-36x+108y+36y^2+144y+144=256.$

$9x^2+117y^2-54xy-36x+252y+180=256.$

$9x^2+117y^2-54xy-36x+252y-76=0.$

Match with the given form

Given locus: $9\left(x^2+\alpha y^2+\beta xy+\gamma x+28y\right)-76=0.$

Expanding: $9x^2+9\alpha y^2+9\beta xy+9\gamma x+252y-76=0.$

Compare with $9x^2+117y^2-54xy-36x+252y-76=0.$

So, $9\alpha=117 \Rightarrow \alpha=13,$ $9\beta=-54 \Rightarrow \beta=-6,$ $9\gamma=-36 \Rightarrow \gamma=-4.$

Compute the required value

$\alpha-\beta-\gamma=13-(-6)-(-4)=13+6+4=23.$

Check with options

The correct option is $\boxed{23}$ which is Option D.