Circle Intersection Points with Coordinate Axes from Matrix Eigenvalues

To find the largest possible value of $\alpha$, we first solve the characteristic equation $\det(A - \alpha I) = 0$ for the matrix: $A = \begin{bmatrix} 1 & 2 & 7 \\ 4 & -2 & 8 \\ 3 & 8 & -7 \end{bmatrix}$

The determinant is given by: $\det(A - \alpha I) = \begin{vmatrix} 1 - \alpha & 2 & 7 \\ 4 & -2 - \alpha & 8 \\ 3 & 8 & -7 - \alpha \end{vmatrix} = 0$

Expanding the determinant along the first row: $(1 - \alpha)[(-2 - \alpha)(-7 - \alpha) - 64] - 2[4(-7 - \alpha) - 24] + 7[32 - 3(-2 - \alpha)] = 0$ $(1 - \alpha)(\alpha^2 + 9\alpha - 50) - 2(-4\alpha - 52) + 7(3\alpha + 38) = 0$ $-\alpha^3 - 8\alpha^2 + 88\alpha + 320 = 0$ $\alpha^3 + 8\alpha^2 - 88\alpha - 320 = 0$

To find the roots, we test $\alpha = 8$: $8^3 + 8(8^2) - 88(8) - 320 = 512 + 512 - 704 - 320 = 0$

Factorizing the cubic equation by dividing by $(\alpha - 8)$: $(\alpha - 8)(\alpha^2 + 16\alpha + 40) = 0$

The roots of $\alpha^2 + 16\alpha + 40 = 0$ are given by the quadratic formula: $\alpha = \frac{-16 \pm \sqrt{256 - 160}}{2} = -8 \pm 2\sqrt{6}$

Thus, the real roots are $\alpha = 8$, $-8 + 2\sqrt{6}$, and $-8 - 2\sqrt{6}$.

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

Now, substituting $p = 8$ into the equation of the circle: $(x - 8)^2 + (y - 16)^2 = 320$

To find the intersection points with the coordinate axes:

1. Intersection with the x-axis ($y = 0$): $(x - 8)^2 + (0 - 16)^2 = 320$ $(x - 8)^2 + 256 = 320$ $(x - 8)^2 = 64 \implies x - 8 = \pm 8$ $x = 0 \quad \text{or} \quad x = 16$ The points of intersection on the x-axis are $(0, 0)$ and $(16, 0)$.

2. Intersection with the y-axis ($x = 0$): $(0 - 8)^2 + (y - 16)^2 = 320$ $64 + (y - 16)^2 = 320$ $(y - 16)^2 = 256 \implies y - 16 = \pm 16$ $y = 0 \quad \text{or} \quad y = 32$ The points of intersection on the y-axis are $(0, 0)$ and $(0, 32)$.

Combining the unique intersection points, we have:

1. $(0,0)$ (the origin)
2. $(16, 0)$
3. $(0, 32)$

Therefore, the circle intersects the coordinate axes at 3 points.

Correct Option: C