A rod of length eight units moves such that its ends A and B always...

$$\begin{aligned} & A B=8 \\ & A B^2=64\end{aligned}$$
$\Rightarrow(a-b)^2+(b+4)^2=64$ ...(1) Now $P$ divides $A B$ in the ratio $2: 1$ internally $\Rightarrow h=\frac{2 a+b}{3}$ and $k=\frac{-4+b+2}{3} \Rightarrow 2 a+b=3 h \ldots$ (2) $k=\frac{b-2}{3}$ From equation (2) and (3) $\quad \Rightarrow b=3 \mathrm{k}+2$$\begin{aligned} & \Rightarrow \quad 2 a=3 h-3 k-2 \\ & \Rightarrow \quad a=\frac{3 h-3 k-2}{2} \end{aligned}$Now by putting value of $a and b$ in equation\begin{aligned} & \Rightarrow\left(\frac{3 h-3 k-2}{2}-(3 k+2)\right)^2+(3 k+2+4)^2=64 \\ & \Rightarrow\left(\frac{3 h-3 k-2-6 k-4}{2}\right)^2+(3 k+6)^2=64 \\ & \Rightarrow(3 h-9 k-6)^2+4(3 k+6)^2=4 \times 64 \\ & \Rightarrow 9(h-3 k-2)^2+36(k+2)^2=256 \\ & \Rightarrow 9\left(h^2+9 k^2+4-6 h k-4 h+12 k\right) \\ & \quad+36\left(k^2+4+4 k\right)=256 \\ & \Rightarrow 9\left(h^2+13 k^2+20-6 h k-4 h+28 k\right)=256 \end{aligned}Replacing $h$ by $x and k$ by $y\begin{aligned} & \Rightarrow 9\left(x^2+13 y^2-6 x y-4 x+28 y\right)+180-256=0 \\ & \Rightarrow 9\left(x^2+13 y^2-6 x y-4 x+28 y\right)-76=0 \end{aligned}$$$By comparing$\alpha=13, \beta=-6, \gamma=-4 \alpha-\beta-\gamma=13+6+4=23$