The equation of the (upside down) cone is:
$$
f(x,y,z) = x^2+y^2-z^2 = 0
$$
Note that the top of the cone is at the origin $(0,0,0)$.
Take another point $(a,b,c)\ne(0,0,0)$ at the surface of the cone:
$$
a^2+b^2-c^2 = 0
$$
The equation of the OP's line in $\mathbb{R}^{3}$ is required to be of the form:
$$
\begin{bmatrix} x \\ y \\ z \end{bmatrix} =
\begin{bmatrix} a \\ b \\ c \end{bmatrix} t
$$
So the vector $(a,b,c)$ should preferrably be normed:
$$
a^2+b^2+c^2 = 1
$$
Two equations with three unknowns; we have one degree of freedom. A suitable solution is:
$$
\begin{cases} a = \cos(\phi)/\sqrt{2} \\ b = \sin(\phi)/\sqrt{2} \\ c = 1/\sqrt{2} \end{cases}
$$
To be interpreted as a vector to any point (angle $\phi$) on the circle $\,x^2+y^2=1/2\,$ at height $\,z=1/\sqrt{2}\,$.
The end-result is:
$$
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} \cos(\phi) \\ \sin(\phi) \\ 1 \end{bmatrix} t
$$
EDIT. Before I forget, this line is, of course, in the tangent plane.
$$
\vec{\nabla} f(x=a,y=b,z=c) = \begin{bmatrix} 2x \\ 2y \\ -2z \end{bmatrix} = 2\begin{bmatrix} a \\ b \\ -c \end{bmatrix}
$$
So the equation of the tangent plane is (norming factors omitted):
$$
ax+by-cz=0 \quad \mbox{with} \quad
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} t
\quad \Longrightarrow \quad (a^2+b^2-c^2)t^2=0
$$
which is the same as saying that $(a,b,c)$ is at the surface of the cone.