They do. the initial point is same in both lines.
formally,
$\vec{r_2}=(2\hat i +3 \hat j +\hat k)+t(2\hat i+2\hat j +\hat k)$
if they intersect then $\begin{vmatrix} 4 & 0 & -1 \\ 2 & 2 & 1 \\ (2-2) & (3-3) & (1-1) \end{vmatrix}=0$
(which obviously is zero)