Identify an arbitrary elliptical cylinder with $9$ points

Let $r = (x, y, z)$ , then the equation of an elliptical cylinder is

$(r - r_0)^T Q (r - r_0) = 1$ , where $r_0$ is the position vector of any point on the axis, and $Q$ can be factored as $Q = R D R^T$ , with $D = \text{diag} \{ d_1, d_2, 0 \}$ , so that the third column of the rotation matrix $R$ is a unit vector along the axis of the cylinder. Expanding the equation of the cylinder, we get,

$r^T Q r - 2 r^T Q r_0 + r_0^T Q r_0 - 1 = 0$

Now, let $h = -2 Q r_0 , c = r_0^T Q r_0 - 1$ , then for each point $r_i = (x_i, y_i, z_i ), i = 1, 2, ..., 9$ , we have

$Q_{11} x_i^2 + Q_{22} y_i^2 + Q_{33} z_i^2 + 2 Q_{12} x_i y_i + 2 Q_{13} x_i z_i + 2 Q_{23} y_i z_i + h_1 x_i + h_2 y_i + h_3 z_i + c = 0$

This is a system of nine equations in ten unknowns, and can be solved using standard linear algebra methods, to yield:

$Q = t Q_0, h = t h_0, c = t , t \in \mathbb{R}$

where

$Q_0 = \begin{bmatrix} 0.029143375 && 0.015755377 && -0.037231244 \\ 0.015755377 && 0.016886522 && -0.011154215 \\ -0.037231244 && -0.011154215 && 0.057185627 \end{bmatrix}$

$h_0 = \begin{bmatrix} -0.548254535 \\-0.463773453 \\ 0.520934247 \end{bmatrix}$

Now since $h = -2 Q r_0$ , then $h_0 = -2 Q_0 r_0$ , and this system can be solved for $r_0$ , and there are infinite solution as one would expect because $r_0$ can be any point on the axis of the cylinder. Solving this $3 \times 3$ system of equations yields:

$r_0 = (4, 10, 0) + s (1.857197472, -1.07225346, 1 ) , s \in \mathbb{R}$ . Taking $s = 0$ , then $x_0 = 4 , y_0 = 10$ are the coordinates of the point of intersection of the axis with the $xy$ plane, and $r_0 = (4, 10, 0)$

Next, diagonalize $Q_0 = R D_0 R^T$ , you get,

$D_0 = \text{diag} \{ 0.015582822 , 0.087632701, 0 \}$

with a corresponding $R = \begin{bmatrix} 0.250013627 && 0.566963696&& 0.784885567 \\ 0.855654565 &&0.250013627&& -0.453153894\\ 0.453153894&& -0.784885567&& 0.422618262\end{bmatrix}$

Since $c = t = r0^T Q r_0 - 1 = r_0^T (t Q_0) r_0 - 1$ , then $t = \dfrac{1 }{r_0^T Q_0 r_0 - 1 } = 0.414014158243576$

Therefore, the diagonal matrix $D = t D_0 = \text{diag} \{ 0.006451509 , 0.036281179, 0 \}$

The semi-major axis length $= a = \dfrac{1}{\sqrt{0.006451509}} = 12.45$

And the semi-minor axis length $= b = \dfrac{1}{\sqrt{0.036281179}} = 5.25$

The vector $u = (0.784885567 , -0.453153894, 0.422618262 )$ is the unit vector along the axis of the cylinder, and

The vector $v = (0.250013627 , 0.855654565, 0.453153894 )$ is the unit vector along the major axis of the elliptical cylinder's cross section.

Hence, $S = 4 + 10 + 12.45 + 5.25 + 0.784885567 + 0.453153894 + 0.422618262 + 0.250013627 + 0.855654565+0.453153894 = 34.91947981$ , so that the answer is $\lfloor 34.91947981 \times 10^5 \rfloor = \boxed{3491947}$