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 $t = -1$?)
Best Answer
There are a few ways to solve this problem.
Method 1.
Method 2.
Write an equation (in the form $f(x)=mx+b$) for the line passing through the given points $P_0$ and $P_1$.
Using the equation from 1, the coordinates for any point on the given line can be written in the form $(x, f(x))$.
Use the distance formula to write a formula for the distance between the given point $P$ and an arbitrary point $(x, f(x))$ on the line. This formula will be a function of $x$.
Find the minimum value of the function from 7. There are a few different ways to do this, too: