I agree it is difficult to interpret the question.
If you write the lines in the more useful $y=mx+c$ form you find
$$L_1:=y=-\frac43 x+3$$
and
$$L_2:=y=-\frac{4}{3}x+1.$$
This means that the lines have the same slopes and so are parallel. Also the first cuts at $y=3$ and the second at $y=1$.
I think the "intercept of length three" means that the line we are looking for, $L$, cuts $L_1$ and $L_2$ at points $P_1=L\cap L_1$ and $P_2=L\cap L_2$ such that
$$|P_1P_2|=3.$$
There are a number of ways of finishing this off. Synthetic geometry is probably easiest.
Remember: Coordinate Geometry is supposed to make Synthetic Geometry
easier! You are perfectly entitled to use Synthetic Geometry to make
Coordinate Geometry easier also!
Consider the point $R$ on $L_1$ that is found vertically above $P_2$ and the triangle $\Delta (RP_1P_2)$.
We know the angle at $R$ (draw a picture and use the slope of $L_1$).
Using the Sine Rule we can thus find the angle at $P_1$ as $|P_2R|=2$ (why?).
Hence you know all the angles in particular the angle at $P_2$, $\alpha$.
The slope of $L$ is thus
$$m=\tan\left(\frac{\pi}{2}-\alpha\right).$$
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
Well, recall the length of a perpendicular from point $(x_{1},y_{1})$ to the line $ax+by+c=0$ is equal to the distance between them and is thus given by $$\dfrac{|ax_{1}+by_{1}+c|}{\sqrt{a^2+b^2}}$$
So the length of a perpendicular from point $(2a,2a)$ on the line $mx-y+a=0$ would be given by $$\dfrac{|2ma-a|}{\sqrt{m^2+1}}=a$$ So squaring and dividing by $a^2$ we have $$\dfrac{4m^2-4m+1}{m^2+1}=1 \implies m=0 \text{ or } m=\dfrac{4}{3}$$