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 $t = -1$?)
The points on the infinite right circular cylinder are at distance $r$ from the axis of the cylinder. The distance between point $\vec{p}$ and line through $\vec{p}_1$ and $\vec{p}_2$ is
$$d = \frac{\left \lvert (\vec{p} - \vec{p}_1) \times (\vec{p} - \vec{p}_2) \right \rvert}{\left \lvert \vec{p}_1 - \vec{p}_2 \right \rvert}$$
so we can square that to get our equation describing an infinite right circular cylinder of radius $r$ whose axis passes through points $\vec{p}_1$ and $\vec{p}_2$:
$$r^2 = \frac{\left \lvert (\vec{p} - \vec{p}_1) \times (\vec{p} - \vec{p}_2) \right \rvert^2}{\left \lvert \vec{p}_1 - \vec{p}_2 \right \rvert^2}$$
That can be written using Cartesian coordinates as
$$r^2 = \frac{ \left ( (y-y_1)(z-z_2) - (z-z_1)(y-y_2) \right )^2 }{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2 } + \frac{ \left ( (z-z_1)(x-x_2) - (x-x_1)(z-z_2) \right )^2 }{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2 } + \frac{ \left ( (x-x_1)(y-y_2) - (y-y_1)(x-x_2) \right )^2 }{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2 }$$
which we can convert to standard form by multiplying by the right-hand-side denominator, and moving all to the same side:
$$\left ( (y-y_1)(z-z_2) - (z-z_1)(y-y_2) \right )^2 + \left ( (z-z_1)(x-x_2) - (x-x_1)(z-z_2) \right )^2 + \left ( (x-x_1)(y-y_2) - (y-y_1)(x-x_2) \right )^2 - r^2 \left ( (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2 \right ) = 0$$
I have at my Wikipedia talk page the equations to locate the intersection between a line passing through origin and an arbitrary right circular cylinder; without caps, with flat end caps, or with spherical end caps. The key point is that when the ray (or line we are testing intersection for) passes through origin (so we use an unit vector $\hat{n}$, $\lvert\hat{n}\rvert=1$, to describe the line or ray), the equations become rather straightforward.
If the underlying problem is related to grinding spheres using a grinding wheel -- a cutting plane, essentially -- then a simplified coordinate system where origin is the sphere origin, and $\hat{z}$ is along the rotation axis, makes for easy calculations. Consider this illustration:
![Grinding sphere]](https://web.archive.org/web/20190515124011/http://www.nominal-animal.net/answers/tilted-sphere.svg)
The contact point scribes a circle around the axis of rotation. We can describe this circle using a single scalar, $d$, which is the signed distance from the origin to the plane of that circle, along the axis of rotation. If the circle is not on the surface of the sphere, but perhaps inside it, then we need a second scalar, $r$, to describe the radius of the circle (around the axis of the rotation).
If the radius of the sphere is $R$, then on the surface of the sphere $r = \sqrt{R^2-d^2}$, of course.
If we include the cutting or abrasion done by the wheel, then two pairs of scalars are needed: $(d_1,r_1)$ and $(d_2,r_2)$. If we assume the grinding wheel is planar, and stays at a fixed location with a perfect cut for at least one rotation of the sphere, the removed section of the sphere leaves a flat facet. Mathematically, $$r(d) = r_1 + \left ( d - d_1 \right ) \frac{r_2 - r_1}{d_2 - d_1}$$
where $d_1 \le d \le d_2$. Remember, $d$ is the distance along the axis of rotation, and $r$ the circle radius (around a point on the axis of rotation, at distance $d$ from the center of the sphere); that is why the dependence above is linear too.
If the axis of the grinding wheel always intersects the rotation axis, and you call up $z$ and right $x$ in the above diagram, then $(d,r) = (z,x)$, and the primary problem (surface of revolution caused by grinding, or shape of the "sphere" after several grinds) reduces to planar (2D) vector calculus.
If we also have the axis of rotation $\hat{n}$ in some fixed sphere coordinate system, it is easy to calculate the "ribbons" each grind cuts into the sphere, either parametrically (using say $0 \le u \le \pi$ along the circular ribbon, and $0 \le v \le 1$ across the ribbon -- useful for visualization) or algebraically.
![Grinding sphere]](http://www.nominal-animal.net/answers/tilted-sphere.svg)
Best Answer
Actually, there are infinitely many points as 1 unit far from the line. For those points, namely having coordinates $(p,q,r)$, there exists a unit vector $(a,b,c)$ normal to the vector of line $(1,1,1)$ and a point $(x,y,z)$ on the line such that $$ (p,q,r)=(x,y,z)+(a,b,c). $$ The unicity of $(a,b,c)$, along its normality implies $$ {a^2+b^2+c^2=1 \\ (a,b,c)\cdot(1,1,1)=a+b+c=0, } $$ which yields $$ {c=-a-b \\ a^2+b^2+ab=\frac{1}{2}\implies \\ b=\frac{-a\pm\sqrt{2-3a^2}}{2} \\ c=\frac{-a\mp\sqrt{2-3a^2}}{2}. } $$Since the point $(x,y,z)$ is arbitrary, the equation of all points far from the line as much as 1 is $$ \left(t+a,t+\frac{-a\pm\sqrt{2-3a^2}}{2},1+t+\frac{-a\mp\sqrt{2-3a^2}}{2}\right) $$ for $|a|\le \sqrt\frac{2}{3}$.