Transform matrix to get intersection point of line and plane Give a point $Q[A, B, C]$ and a plane $Dx + Ey + Fz = 0$.
Now for any point $P[x, y, z]$, is there a matrix $M$ that can be used to get the intersection point of the line $PQ$ and the plane
(Say intersection point is $F, P \times M = F$)?
 A: Writing $n=[D,E,F]$ for the normal of the plane, you are looking for a $\lambda$ such that $$\langle n,Q+\lambda (P-Q)\rangle=0\,,$$ i.e. the scalar product between the normal and a generic point on the line equals zero at the point of intersection. Rewrite that to
\begin{align*}
\langle n,Q\rangle+\lambda\langle n,P-Q\rangle&=0\\
\lambda &= \frac{\langle n,Q\rangle}{\langle n,Q-P\rangle}
\end{align*}
At this point you see that $P$ is in the denominator; you are dividing by some quantity which depends on $P$. Expressing that as a matrix multiplication in affine geometry is a bit tricky.
There is however an approach which deals with divisions in geometry more elegantly, namely projective geometry. It uses homogeneous coordinates, so now you have $n=[D:E:F:0]$, $Q=[A:B:C:1]$ and $P=[x:y:z:1]$. Any multiple of any of these vectors describes the same object. This time a generic point on the line is any linear combination, so you have
\begin{align*}
\langle n, \lambda P+\mu Q\rangle &= 0 \\
\lambda\langle n,P\rangle+\mu\langle n,Q\rangle &= 0
\end{align*}
Here you have a one-dimensional space of solutions, and any one will represent the same geometric point of intersection. Pick $\lambda=\langle n,Q\rangle$ and $\mu=-\langle n,P\rangle$. The point of intersection was called $F$ in your question, but as that conflicts with one parameter of your plane, let's use $R$ instead.
$$R=\langle n,Q\rangle P-\langle n,P\rangle Q$$
If you treat your vectors as line vectors (as your use of notation suggests), you can write this as
$$R=n\cdot Q^T\cdot P - n\cdot P^T\cdot Q$$
as the scalar product is the product of a row vector with a column vector (the transpose of a row vector). Now we try to get the $P$ to the very left.
\begin{align*}
R &= P\cdot n\cdot Q^T - P\cdot n^T\cdot Q \\
R &= P\cdot\mathbf 1_4\cdot n\cdot Q^T - P\cdot n^T\cdot Q \\
R &= P\cdot\underbrace{\left(n\cdot Q^T\cdot\mathbf 1_4 - n^T\cdot Q\right)}_M
\end{align*}
The $n^T\cdot Q$ there is an outer product: column vector times row vector. It results in a $4\times4$ matrix. The left term $n\cdot Q^T$ is just a regular scalar product. To make that compatible for subtraction of a matrix, we multiply with the $4\times 4$ unit matrix $\mathbf 1_4$.
Going back to your notation you get that matrix $M$ as
$$M=\begin{bmatrix}
BE+CF & -BD & -CD & -D \\
-AE & AD+CF & -CE & -E \\
-AF & -BF & AD+BE & -F \\
0 & 0 & 0 & AD+BE+CF
\end{bmatrix}$$
With this matrix you get $R=P\cdot M$ as the homogeneous coordinates of the point of intersection. Divide by the last coordinate to obtain $x,y,z$ coordinates in the first three coordinates.
One nice benefit of doing all of this projectively: if your plane is not through the origin, it still fits this picture: $n=[D,E,F,G]$ would describe a plane $Dx+Ey+Fz+G=0$.
