Equation for a line through a plane in homogeneous coordinates. For calculations in 2D space, there exist a few useful equations to compute general geometry with the vector dot product . and the vector cross product x when working with homogeneous coordinates (remember even though we are working in 2D space the homogeneous representation of lines and point are 3D vectors):
1) The point where two lines cross x = l1 x l2.
2) The line between two points l = x1 x x2
3) A point lies on a line if x . l = 0
In 3D space a plane can be described by a normal vector n = [n1, n2, n3] and a point on the plane x = [x1, x2, x3] both in Euclidean coordinates.
In homogeneous coordinated the plane can be defined as p = [n1, n2, n3, -(n1*x1 + n2*x2 + n3*x3)]. 
Is there a short equation for finding the point where a line passes through a plane? Intuitively it feels as it should be x = l x p in homogeneous coordinates, but this computation does not exist, since there is no cross product in 4 dimensions.
At the moment I am only able to compute the intersection by defining the line with the equation l(t) = a + b(t), where a is a point on the line and b is the direction of the line in Euclidean coordinates.
For a = [a1, a2, a3], b = [b1, b2, b3] and the plane in question p = [p1, p2, p3, p4], the point of intersection x = [x1, x2, x3]  can be obtained by substituting t in the line equation with t = -(p . [a1, a2, a3, 1])/(p . [b1, b2, b3, 0])
In summary, is there a elegant equation to find a point x where a line l crosses through a plane p preferably in homogeneous coordinates?
 A: Definition
You are right, 3D points and planes are described with 4 homogeneous coordinates. A point at $\vec{r}$ is $P=(\vec{r};1)$ and a plane $W=(\vec{n};-d)=(\vec{n};-\vec{r}\cdot\vec{n})$ with normal $\vec{n}$ through point $\vec{r}$, or with minimum distance to origin $d$.
A line needs 6 coordinates (plücker coordinates) describing the direction and moment about the axis. A line along $\vec{e}$ through a point $\vec{r}$ has coordinates $L=[\vec{e};\vec{r}\times\vec{e}]$. Given a line $L=[\vec{l};\vec{m}]$, the direction is recovered by $\vec{e}=\frac{\vec{l}}{|\vec{l}|}$ and the position by $\vec{r} = \frac{\vec{l}\times\vec{m}}{|\vec{l}|^2}$
Now derive the point $P=(\vec{r};1)$ where line $L=[\vec{l};\vec{m}]$ meets plane $W=(\vec{w};\epsilon)$ as follows:


*

*See that for the point to be on the plane you must have $\epsilon = - \vec{r}\cdot \vec{w}$

*For the point to be on the line you must have $\vec{m} = \vec{r} \times \vec{l}$

*Use the vector triple product to get
$$ 
  \vec{w} \times \vec{m} = \vec{w} \times \left( \vec{r} \times \vec{l} \right) = \vec{r} (\vec{w}\cdot\vec{l})-\vec{l}(\vec{w}\cdot\vec{r}) $$


$$ \vec{w} \times \vec{m} = \vec{r} (\vec{w}\cdot\vec{l}) - \vec{l}(-\epsilon) $$
$$ \vec{r} = \frac{\vec{w}\times\vec{m}-\epsilon \vec{l}}{\vec{w}\cdot\vec{l}} $$


*

*Define the line-plane meet operator as


$$ \begin{aligned}
 P & = [W\times] L \\
 \begin{pmatrix} \vec{p} \\ \delta \end{pmatrix} & = \begin{bmatrix} 
-\epsilon {\bf 1} & \vec{w}\times \\ \vec{w}^\top & 0
\end{bmatrix} \begin{pmatrix} \vec{l} \\ \vec{m} \end{pmatrix}
\end{aligned}$$
where $\vec{w}\times = \begin{pmatrix}x\\y\\z\end{pmatrix} \times = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}$ is the cross product matrix operator in 3×3 form.


*

*The meet operator $[W\times]$ has dimensions 4×6 to work between lines and points.



Example


*

*A plane normal to the $x$ axis located at $x=3$ has coordinates $W=(1,0,0;-3)$

*A line through $y=2$ directed towards $\hat{i}+\hat{k}$ has coordinates $L=[1,0,1;2,0,-2]$

*The meet operator is $$ [W\times] = \left[ \begin{array}{ccc|ccc} 
3 & 0 & 0 & 0 & 0 & 0 \\ 
0 & 3 & 0 & 0 & 0 & -1 \\
0 & 0 & 3 & 0 & 1 & 0 \\
\hline 1 & 0 & 0 & 0 & 0 & 0
\end{array}\right]$$

*The point where the line meets the plane is $P=[W\times]L$ $$P=\left[ \begin{array}{ccc|ccc} 
3 & 0 & 0 & 0 & 0 & 0 \\ 
0 & 3 & 0 & 0 & 0 & -1 \\
0 & 0 & 3 & 0 & 1 & 0 \\
\hline 1 & 0 & 0 & 0 & 0 & 0
\end{array}\right] \begin{bmatrix}1\\0\\1\\ \hline 2 \\ 0 \\ -2  \end{bmatrix} =\begin{pmatrix}3\\2\\3\\ \hline 1 \end{pmatrix}$$

*The point is located at $\vec{r} = (3,2,3)$

A: To answer your question about the line and plane in $\mathbb R^3$, think of it this way: You want to find the line of intersection of a $2$-dimensional subspace $L$ and a $3$-dimensional subspace $P$ in $\mathbb R^4$. Let $\mathbf n$ be the normal vector of $P$; you want the line in $L$ orthogonal to $\mathbf n$. This is really the same as your approach. 
If you want something more like your approach one dimension lower, you need two vectors orthogonal to $L$, i.e., a basis for $L^\perp$, say $\mathbf v_1$ and $\mathbf v_2$. Then compute the cross product $\mathbf n\times \mathbf v_1\times \mathbf v_2$, and, voilà! You compute the cross product of $n$ vectors in $\mathbb R^{n+1}$ exactly as in the case of $n=2$.
A: In a similar vein to ja72’s answer, you can use the Plücker matrix of a line to compute its intersection with a plane: The Plücker matrix of a line through points $\mathbf p$ and $\mathbf q$ is $L=\mathbf p\mathbf q^T-\mathbf q\mathbf p^T$, and its intersection with the plane $\mathbf\pi$ is simply $$L\pi = (\mathbf q^T\mathbf\pi)\mathbf p - (\mathbf p^T\mathbf\pi)\mathbf q.$$ If this product vanishes, then the line lies on $\mathbf\pi$. The Plücker matrix $L$ is in a certain sense the three-dimensional analog of the formula $\mathbf l = \mathbf p\times\mathbf q$ in 2-D.
