$\left| \begin{matrix} 1-\alpha & 2 & 7\\ 4 & -2-\alpha & 8\\ 3 & 8 & -7-\alpha \end{matrix} \right| =0$

$\Rightarrow (1-\alpha)\left[(\alpha+2)(\alpha+7)-64\right] -2\left[-28-4\alpha-24\right] +7\left[32+6+3\alpha\right] =0$

$\Rightarrow \alpha^3+8\alpha^2-88\alpha-320=0$

$(\alpha -8)({\alpha }^2+16\alpha +40)=0$

$\alpha =8,\alpha =-8\pm 2\sqrt{6}$

So $p=-8$

$\Rightarrow (x-8)^2+(y-16)^2 = 320 = 8^2+16^2$

put $y=0 \Rightarrow x=16,\;0$

put $x=0 \Rightarrow y=32,\;0$

$\Rightarrow (16,0),\;(0,32),\;(0,0)$

$\Rightarrow 3$ points of intersection