First find the pole (point) of the line ${\bf L}=\pmatrix{a \\ b \\ c}$ using the conic $\mathtt{C}=\left[ \matrix{A & C & D \\ C & B & E \\ D & E& F} \right]$. This is found using the inverse of the conic
$$ \mathbf{P} = \mathtt{C}^{-1} \mathbf{L} = \pmatrix{u \\ v \\ w}$$
Since we will use the inverse again, set
$$\mathtt{C}^{-1}=\left[ \matrix{a & c & d \\ c & b & e \\ d & e& f} \right]$$
The pencil of lines through $\bf P$ is parametrically defined as
$$\mathbf{T}(\psi) = \pmatrix{-w \sin \psi \\ w \cos \psi \\ u \sin \psi - v \cos \psi}$$
Now if you find a line $\mathbf{T}$ that is tangent to the conic, it will have $\mathbf{T}^\top \mathtt{C}^{-1} \mathbf{T}=0$ and the tangent point $\mathbf{Q}=\mathtt{C}^{-1} \mathbf{T}$ lies on $\mathbf{L}$. So $\mathbf{Q}$ is an intersection point.
To solve $\mathbf{T}^\top \mathtt{C}^{-1} \mathbf{T}=0$ for $\psi$ involves solving an equation of the form $$K_0 + K_1 \sin(2 \psi) + K_2 \cos(2 \psi)=0$$ with $$\begin{align} K_0 & = a w^2-2 d u w + f u^2 \\ K_1 & = -2 ( c w^2-w (d v+e u)+f u v)\\ K_2 &= w^2 (b-a)+2 w (d u-e v) + f (v^2-u^2) \end{align}$$
There are two solutions to the above trig equation
$$ \psi = \begin{cases}
\frac{1}{2} \left( \tan^{-1} \left( \frac{K_1}{K_2} \right) - \sin^{-1} \left( \frac{2 K_0 + K_2}{\sqrt{K_1^2+K_2^2}} \right) -\frac{\pi}{2}\right) & \mbox{solution 1}\\
\frac{1}{2} \left( \tan^{-1} \left( \frac{K_1}{K_2} \right) + \sin^{-1} \left( \frac{2 K_0 + K_2}{\sqrt{K_1^2+K_2^2}} \right) +\frac{\pi}{2}\right) & \mbox{solution 2}
\end{cases} $$
In the end the two intersection points are defined by $\mathbf{Q} = \mathtt{C}^{-1} \mathbf{T}(\psi) $
$$ \mathbf{Q} = \pmatrix{
(c w-d v) \cos \psi + (d u-a w) \sin \psi \\
(b w-e v) \cos \psi + (e u-c w) \sin \psi \\
(e w-f v) \cos \psi + (f u-d w) \sin \psi } $$
Example
Conic $\mathtt{C} = \left[ \matrix{1 & -\tfrac{7}{6} & -3 \\ -\tfrac{7}{6} & 4 & 1 \\ -3 & 1 & \tfrac{13}{6} } \right] $ and line $\mathbf{L} = \pmatrix{3 \\ -2 \\ -7}$. The polar point is $$\mathbf{P} = \mathtt{C}^{-1} \mathbf{L} = \pmatrix{u \\ v \\ w} = \pmatrix{ \tfrac{11208}{5245} \\ \tfrac{1134}{5245} \\ -\tfrac{390}{1049} } $$
The tangent lines are thus $$\mathbf{T}(\psi) = \pmatrix{ \tfrac{390}{1049} \sin\psi \\ -\tfrac{390}{1049} \cos\psi \\ \tfrac{11208}{5245} \sin\psi - \frac{1134}{5245} \cos\psi }$$
The tangency equation $\mathbf{T}^\top \mathtt{C}^{-1} \mathbf{T}=0$ simplifis to the following:
$$ 1.26602894762909 \cos^2 \psi+0.330351717237625 \cos\psi \sin \psi-1.24876309636214=0 $$
with solution
$$ \begin{cases} \psi = 0.299923260411840 & \mbox{solution 1} \\ \psi =-0.0446792696983973 & \mbox{solution 2} \end{cases} $$
The intersection points are
$$ \begin{align} \mathbf{Q} &= \pmatrix{ -0.231100412938826 \\ -0.141550789771816 \\-0.0585999513246922} & \mathbf{Q} &= \pmatrix{ 0.136962490550640 \\-0.0727785333232919 \\0.0794920768997864 } \end{align} $$
Best Answer
A conic curve, is the intersection of the 3D cone $p_1z^2=p_2x^2+p_3y^2$ with the 3D plane $p_4x+p_5y+p_6=0$, which takes the general form as stated in the OP. Now, a point can also be regarded as a conic curve (or conic, simply) in a sense that it is the intersection of the 3D plane with the vertex of the 3D cone, i.e. the point in which, the two halves of the cone meet. An example is the intersection of $z^2=x^2+y^2$ with $2z=x+y$. Hence, your approach is correct. Another way for justifying the result, is to note that any single point is an ellipse with zero diameters and ellipse is a conic curve.