Since no three lines are concurrent, there are $ \binom{n}{2} $ points in the plane, and each line consists of $ (n-1) $ points, as no three lines are concurrent (i.e. This plane has maximum possible no. of points in a plane of $n$ non-parallel lines). Since $ n > 2 \Rightarrow n \ge 3$, we can restore the diagram by making lines that pass through many different possible $ n-1 $ points.

The answer is $ \boxed{Yes!} $. Now we prove that we can restore the diagram... CLAIM: There are $ (n-1) $ points on any line. The key observation is that no three lines are concurrent, which means that this is plane of $ n $ lines with the maximum no. of intersections, and further there are $ \binom{n}{2} $ points in the plane. Hence, there are $ (n-1) $ points on any line. CLAIM: The diagram can be restored. From the hypothesis $ n > 2 $, there is exactly one distinct line that passes through $ n-1 $ points. Hence, the diagram can be restored.

We claim the answer is yes. This is achieved by greedily drawing lines between $n-1$ collinear intersection points. Indeed, given any $n-1$ intersection points, you can clearly find a pair of points on the same line since $2(n-1) > n$. So, if $n-1$ points are collinear they must all be on one of our original lines as desired.