Find the locus of a fixed-ratio point between a line and a plane

$l:\mathbf r=(1,-2,1)+\lambda(1,-1,2)$; $p_1:\mathbf r=(0,0,12)+\alpha(1,-1,2)+\beta(0,1,-5)$. The direction $(1,-1,2)$ is parallel to both, so the line is parallel to the plane. It is not contained because its point has normal constant −6 rather than 12.

A normal is $(1,-1,2)\times(0,1,-5)=(3,5,1)$. Its dot product with the line direction is $3-5+2=0$, again proving the parallel direction.

$OA=(1,-2,1)+\lambda(1,-1,2)$ and $OB=(0,0,12)+\alpha(1,-1,2)+\beta(0,1,-5)$. Since BC=2AC,$OC=(OB+2OA)/3$. Thus

$\begin{aligned}&OC\\ & =\frac13(2,-4,14)\\ &{}+\frac{\alpha+2\lambda}{3}(1,-1,2)\\ &{}+\frac\beta3(0,1,-5).\end{aligned}$

The last two coefficients can vary independently over all real numbers, giving a plane through $(2,-4,14)/3$ parallel to those two nonparallel directions.

The same two directions give normal $(3,5,1)$. Hence $p_2:\mathbf r\cdot(3,5,1)=\frac13(2,-4,14)\cdot(3,5,1)=\frac13(6-20+14)=0$, as required.

Use the line point $(1,-2,1)$: distance=$|(1,-2,1)\cdot(3,5,1)|/\sqrt{9+25+1}=|3-10+1|/\sqrt{35}=6/\sqrt{35}$.

p1,p2 are parallel and the section ratio places p2 one third of the way from the line’s parallel level toward p1. Therefore the p1-to-p2 distance is twice the l-to-p2 distance:$2(6/\sqrt{35})=12/\sqrt{35}$.

p1 has scalar-product equation $\mathbf r\cdot(3,5,1)=(0,0,12)\cdot(3,5,1)=12$. Therefore its distance from p2 is $|12-0|/\sqrt{35}=12/\sqrt{35}$.