Find equations of the line through a given point that meets another line at certain distance 
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? 
 A: For a parametric equation of a line $(x,y,z) = (x_0, y_0, z_0) + (a,b,c)s$, the line itself is the set of all points $P(x,y,z)$ such that $x = x_0 + as$, $y = y_0 + bs$, and $z = z_0 + cs$. When $s = 1$, this gives you one point $A$; when $s = 2$, this gives you another point $B$; and when you take all values of $s \in \mathbb{R}$, you get the entire line.
Notice that if instead you had $(x,y,z) = (x_0, y_0, z_0) + (a,b,c)s/2$, you'd still have the same line, except this time, $s$ has to be $2$ to give you point $A$, $s = 4$ gives you point $B$, etc. So when you're trying to find the value of $s$ for any one point, you can just choose any $s$ that you want!
So if $(2,1,1) = (a,b,c)s$, choose any $s$ you want and solve it for $a$, $b$, and $c$.
(By the way, there are two points on the line $(1,2,0) + (2,-1,2)t$ that are $3$ units away from $(1,2,0)$. You found one when you set $t = 1$; what if you set $t = -1$?)
A: 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. 
