Find the vector and parametric equation of the line through point $P(1,0,1)$ that meets the line with vector equation $$p = (1,2,0) + (2,-1,2)t$$ at points a distance $3$ from $P_0(1,2,0)$
How am I supposed to find a point a distance $3$ from $P_0(1,2,0)$? I get the equation of a sphere: $9 = (x-1)^2 + (y-2)^2 + z^2$, where $P(x,y,z)$ is any point 3 units away. This gives infinite points, but to get the one on the line $p$ I did $$\|P_0P\| = t\|v\|,\\ 3 = t\sqrt{9},\\ t = 1$$ This gives $P(3,1,2)$ then the other line passing through those two points is $$(3,1,2) = (1,0,1) + (a,b,c)s , \\ (2,1,1) = (a,b,c)s$$ this gives three equations $2 = as$, $1 = bs$, $1 = cs$. What do I do now? How can I find a,b,c?
First find the point(s) of the line $r : p=(1,2,0)+(2,-1,2)t$ whose distance to $P_0$ equals $3$. Now find the equation of the line that goes through $P$ and the point you found in the first part.