I'll work through my current progress until I reach the bit where I get stuck.
We are given $$n = (1, 0, -2)$$
Thus Cartesian form will be $$x - 2z = r\cdot n$$
Now, $$r\cdot n = (1, -1, -3)\cdot (1, 0, -2) = 7$$
Therefore the Cartesian form of the line is $$x - 2z = 7$$
And here is where I get stuck. I'm not sure how to go from here to get the vector form. All I know is that the vector form has the general form: $$r = r_0 +su + tv$$
To find the vector form, we start with the Cartesian form given. It needs to be solved like a system of linear equations (which, essentially, it is).
We can start with an augmented matrix: $$ \left[ \begin{array}{ccc|c} 1&0&2&7\\ \end{array} \right] $$
We note that it's already in reduced row echelon form. Thus, we can begin solving it. We're going to need to use parameters, since we are solving 1 equation for 3 variables. So, let:
$$y = s$$ $$z = t $$ Where both t and s are members of the real numbers.
Clearly, $$x = 7 + 2z$$
So, using the general form of the vector form given in the question, we can write $$(x, y, z)=(7+2t, s, t) = r_0 + su+tv$$
If we make the parameters equal to zero, we can get $$r_0 = (7, 0 ,0)$$
To find u and v, we simply look at the vector $(7 + 2t, s, t)$ and see how many times s and t turn up. So, for $u$, we note that s appears zero times in the first position of the vector, once in the second position of the vector, and zero times in the third position. Thus $u = (0, 1, 0)$. We do the same thing for $v$
So therefore, the vector form of the equation is
$$(x, y, z)=(7+2t, s, t) = (7, 0, 0) + s(0,1,0)+t(2, 0, 1)$$
Hope this makes sense