Let $A=\begin{bmatrix} 1 & 2 & 7 \\ 4 & -2 & 8 \\ 3 & 8 & -7 \end{bmatrix}$ and $\det(A-\alpha I)=0$, where $\alpha$ is a real number. If the largest possible value of $\alpha$ is $p$, then the circle $(x-p)^2+(y-2p)^2=320$, intersects the co-ordinate axes at

The characteristic equation is given by $\det(A-\alpha I) = 0$.

$\det \begin{bmatrix} 1-\alpha & 2 & 7 \\ 4 & -2-\alpha & 8 \\ 3 & 8 & -7-\alpha \end{bmatrix} = 0$

Expanding the determinant along the first row:

$(1-\alpha)[(-2-\alpha)(-7-\alpha) - 64] - 2[4(-7-\alpha) - 24] + 7[32 - 3(-2-\alpha)] = 0$

$(1-\alpha)(\alpha^2 + 9\alpha - 50) - 2(-4\alpha - 52) + 7(3\alpha + 38) = 0$

$-\alpha^3 - 8\alpha^2 + 59\alpha - 50 + 8\alpha + 104 + 21\alpha + 266 = 0$

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

By inspection, $\alpha = 8$ is a root since $8^3 + 8(8^2) - 88(8) - 320 = 512 + 512 - 704 - 320 = 0$.

Factoring out $(\alpha - 8)$, we get:

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

The roots are $\alpha = 8$ and $\alpha = \dfrac{-16 \pm \sqrt{256 - 160}}{2} = -8 \pm 2\sqrt{6}$.

The largest possible value of $\alpha$ is $p = 8$.

Substituting $p = 8$ into the given equation of the circle:

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

To find the intersection points with the x-axis, substitute $y = 0$:

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

$(x-8)^2 + 256 = 320 \Rightarrow (x-8)^2 = 64 \Rightarrow x-8 = \pm 8$

$x = 0$ or $x = 16$

The circle intersects the x-axis at $(0, 0)$ and $(16, 0)$.

To find the intersection points with the y-axis, substitute $x = 0$:

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

$64 + (y-16)^2 = 320 \Rightarrow (y-16)^2 = 256 \Rightarrow y-16 = \pm 16$

$y = 0$ or $y = 32$

The circle intersects the y-axis at $(0, 0)$ and $(0, 32)$.

The distinct points of intersection with the coordinate axes are $(0, 0)$, $(16, 0)$, and $(0, 32)$.

Thus, the circle intersects the coordinate axes at exactly $3$ points.

Answer: $3$ points