Quadrics and lines

Question

Having a quadratic equation like this : Q= $$ \left\{ \begin{array}{c} x^2+y^2-3z^2-4x+6z=0 \\ \\ (x,y,z)^T \in \mathbb R^3 \end{array} \right. $$ How can I write a cartesian equation of a line on Q and a line on Q passing from the origin ?

Answer 1

Let $(at,bt,ct)$ be a point on the line.

Thus, for all real $t$ we have: $$a^2t^2+b^2t^2-3c^2t^2-4at+6ct=0,$$ which gives $$a^2+b^2=3c^2$$ and $$-4a+6c=0.$$ Can you end it now?

I got two these lines: $$(3t,\sqrt3t,2t)$$ and $$(3t,-\sqrt3t,2t)$$