For the line vec r×vec p=vec q and the plane vec r·vec N=α…
For the line $\vec r\times\vec p=\vec q$ and the plane $\vec r\cdot\vec N=\alpha$, prove that the piercing point is $\vec r=\frac{\alpha\vec p+\vec N\times\vec q}{\vec p\cdot\vec N}$.
Explanation of Method
Proof of the vector formula for the point where a line pierces a plane.
Step 1: Short proof
Let $\vec r$ be the piercing point, so $\vec r\times\vec p=\vec q$ and $\vec r\cdot\vec N=\alpha$. Form $\vec N\times\vec q$ and expand with the rule $\vec a\times(\vec b\times\vec c)=\vec b(\vec a\cdot\vec c)-\vec c(\vec a\cdot\vec b)$:
\[ \vec N\times\vec q=\vec N\times(\vec r\times\vec p)=\vec r\,(\vec N\cdot\vec p)-\vec p\,(\vec N\cdot\vec r)=(\vec p\cdot\vec N)\,\vec r-\alpha\,\vec p . \]
If $\vec p\cdot\vec N\ne0$, i.e. the line is not parallel to the plane: $\vec r=\frac{\alpha\vec p+\vec N\times\vec q}{\vec p\cdot\vec N}\qquad\blacksquare$.
Conversely, this $\vec r$ indeed lies on the line and in the plane. For $\vec p\cdot\vec N=0$ there is no unique piercing point: the line is parallel to the plane or lies in it.
Result
\[\boxed{\,\vec r=\frac{\alpha\vec p+\vec N\times\vec q}{\vec p\cdot\vec N}\qquad\blacksquare\,}\]
Remark
The source's proof takes a long detour through $\vec r_0=\tfrac{\vec p\times\vec q}{|\vec p|^2}$ and contains unfinished lines (“$\vec p\times\vec q=\vec r_0\times\ldots$”). It also has a sign that it only “adjusts”: identity (7) as used there actually gives $-|\vec p|^2(\vec N\times\vec q)$. The one-line argument above avoids all of this.