Let us assume I have a rotated ellipsoid with its orthogonal main axes given by the three vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ in Cartesian coordinates. It is centered around the origin $(0, 0, 0)^T$. I want to find a simple solution on how to get the 3 (or 6) points $p_x, p_y, p_z$ for which this ellipsoid intersects with the Cartesian coordinate system of base \begin{align} \mathbf{e}_x=(1, 0, 0)^T, \mathbf{e}_y=(0, 1, 0)^T, \mathbf{e}_y=(0, 0, 1)^T. \end{align}
I tried finding these points based on the ellipsoid's rotation angles, but this seems too bloated/complicated and it feels like I missed something important. If someone knows a simple solution to this problem (e.g. based on projections and the vector norms) I would highly appreciate it!
Let $A$ be the matrix having vectors $\mathbf{v}_1/v_1^2$, $\mathbf{v}_2/v_2^2$, $\mathbf{v}_3/v_3^2$ as columns. Then the equation of the ellipsoid can be written as $$ \mathbf{x}^TAA^T\mathbf{x}=1. $$ To find the intersection with $x$ axis substitute there $\mathbf{x}^T=(x,0,0)$ and solve for $x$, and so on for $y$ and $z$ axes. You'll find for instance: $$ x^2=\left({v_{1x}^2\over v_1^4}+{v_{2x}^2\over v_2^4}+{v_{3x}^2\over v_3^4}\right)^{-1}. $$