For two lines to be parallel, the vectors defining their slopes have to be parallel, which means their vectors have to be constant multiples of one another. It's easy to see by inspection that $\langle1,1,0 \rangle$ is not parallel to $\langle1,-2,1\rangle$. This shows that the lines aren't parallel. To prove they don't intersect, set the two lines' equations equal to one another. This gives
$$
\langle3,-1,2\rangle + t\langle1,1,0 \rangle = \langle0,5,2\rangle+s\langle1,-2,1\rangle
$$
Solving this component-wise leads to the system
$$
\begin{align}
3&=s-t\\
-6&=-2s-t\\
0&=s
\end{align},
$$
which is clearly not possible because substituting $s=0$ as given by the last equation makes the first two equations $t=-3$ and $t=6$, which is a contradiction, so the system has no solution, so the lines don't intersect.
For the third part, the distance, there are two simple ways to solve the problem. First, you can use the distance formula and then minimize it. Let $D(s,t)$, where $s$ is the parameter for $l_2$ and $t$ is the parameter for $l_1$, be the square of the distance between arbitrary points on $l_1$ and $l_2$. Then minimizing
$$ D(s,t) = ((3+t)-s)^2 + ((-1+t)-(5-2s))^2 + (2-(2+s))^2$$
gives the minimum distance between the two lines.
Alternatively, and much more easily, you can find a vector $\vec n$ that is perpendicular to both $l_1$ and $l_2$. Then, the distance between $l_1$ and $l_2$ is just the projection of $\overrightarrow{Q_1Q_2}$ onto $\vec n$, where $Q_1$ is a point on $l_1$ and $Q_2$ is a point on $l_2$. You can find $\vec n$ by taking the cross product of $\vec{v_1} = \langle1,1,0 \rangle$ and $\vec{v_2} = \langle1,-2,1\rangle$. That is,
$$
\vec n = \vec v_1 \times \vec v_2.
$$
Using that value, the minimum distance between the lines is
$$
\mbox {proj}_\overrightarrow{Q_1Q_2}\vec n = \frac{\|\overrightarrow{Q_1Q_2}\cdot\vec n\|}{\|\vec n\|}
$$
Best Answer
Two non-parallel lines $p_1+\mathbb R v_1$ and $p_2+\mathbb R v_2$ intersect if and only if $(v_1\times v_2)\cdot(p_1-p_2)=0$.
(But if you're implementing this in floating-point arithmetic, you're going to need to build in some safety margins anyway.)