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 coordinate axes at :

The eigenvalues of $$A$$ are the roots of its characteristic equation $$\det(A-\alpha I)=0$$.

Write $$A-\alpha I = \begin{bmatrix}1-\alpha & 2 & 7\\4 & -2-\alpha & 8\\3 & 8 & -7-\alpha\end{bmatrix}$$ and expand the determinant along the first row:

$$\det(A-\alpha I)= (1-\alpha)\bigl[(-2-\alpha)(-7-\alpha)-8\cdot8\bigr] - 2\bigl[4(-7-\alpha)-8\cdot3\bigr] + 7\bigl[4\cdot8-(-2-\alpha)3\bigr]$$

Simplifying each bracket:

$$(1-\alpha)\bigl[(\alpha^2+9\alpha+14)-64\bigr] = (1-\alpha)(\alpha^2+9\alpha-50)$$
$$-\,2[-28-4\alpha-24] = 104+8\alpha$$
$$+\,7[32+6+3\alpha] = 266+21\alpha$$

Combining,

$$(1-\alpha)(\alpha^2+9\alpha-50)+104+8\alpha+266+21\alpha$$ $$= -\alpha^3-8\alpha^2+59\alpha-50+104+8\alpha+266+21\alpha$$ $$= -\alpha^3-8\alpha^2+88\alpha+320$$

Multiplying by $$-1$$ gives the characteristic polynomial

$$\alpha^3+8\alpha^2-88\alpha-320=0\,,\qquad -(1)$$

Try integral factors of $$320$$. Substituting $$\alpha=8$$ in $$(1)$$ yields

$$8^3+8\cdot8^2-88\cdot8-320 = 512+512-704-320 = 0$$

Hence $$(\alpha-8)$$ is a factor. Divide $$(1)$$ by $$(\alpha-8)$$:

$$\alpha^3+8\alpha^2-88\alpha-320=(\alpha-8)(\alpha^2+16\alpha+40)$$

The quadratic factor gives

$$\alpha = \frac{-16\pm\sqrt{16^2-4\cdot40}}{2} = -8\pm2\sqrt6$$

Thus the three eigenvalues are $$8,\,-8+2\sqrt6,\,-8-2\sqrt6$$. The largest eigenvalue is $$p=8$$.

Now substitute $$p=8$$ into the circle equation:

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

Intersection with the x-axis (put $$y=0$$):

$$(x-8)^2+256 = 320\;\Longrightarrow\;(x-8)^2 = 64\;\Longrightarrow\;x-8=\pm8$$ $$\Rightarrow\;x = 16,\;0$$

Hence the points are $$(16,0)$$ and $$(0,0)$$.

Intersection with the y-axis (put $$x=0$$):

$$64+(y-16)^2 = 320\;\Longrightarrow\;(y-16)^2 = 256\;\Longrightarrow\;y-16=\pm16$$ $$\Rightarrow\;y = 32,\;0$$

Hence the points are $$(0,32)$$ and $$(0,0)$$.

The distinct points of intersection with the coordinate axes are therefore

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

which is a total of $$3$$ points.

Option C which is: $$3$$ points