Find the two points where the shortest distance occurs on two lines

Find the point P on $\vec{AB}$ and point Q on $\vec{CD}$ such that $\vec{PQ}$ is the shortest distance between the lines AB and CD, given $\vec{AB} = \begin{pmatrix} 1\\ 0\\ 2\\ \end{pmatrix} + u\begin{pmatrix} -2\\ 2\\ 1\\ \end{pmatrix} ,\vec{CD} = \begin{pmatrix} 0\\ 1\\ 1\\ \end{pmatrix} + v\begin{pmatrix} 2\\ -1\\ -2\\ \end{pmatrix}$ the normal vector $n=\begin{pmatrix} -3\\ -2\\ -2\\ \end{pmatrix}$

and the shortest distance is $\frac{3}{\sqrt{17}}$. I figured all this out from 4 given points but don't know how to find points P and Q. Please help, I'm stuck...

Once you know that the normal vector is $n$, the vector equation $$\begin{pmatrix} 1\\ 0\\ 2\\ \end{pmatrix} + u\begin{pmatrix} -2\\ 2\\ 1\\ \end{pmatrix} +w\begin{pmatrix} -3\\ -2\\ -2\\ \end{pmatrix} = \begin{pmatrix} 0\\ 1\\ 1\\ \end{pmatrix} + v\begin{pmatrix} 2\\ -1\\ -2\\ \end{pmatrix}$$ is equivalent to a system of three linear equations in three unknowns, which indeed has the unique solution $$u=\frac{19}{17},v=-\frac{15}{17},w=\frac{3}{17}.$$ That tells you that the points on $\vec{AB}$ and $\vec{CD}$ are $$\begin{pmatrix} 1\\ 0\\ 2\\ \end{pmatrix} + \frac{19}{17}\begin{pmatrix} -2\\ 2\\ 1\\ \end{pmatrix} \quad\text{and}\quad\begin{pmatrix} 0\\ 1\\ 1\\ \end{pmatrix} -\frac{15}{17}\begin{pmatrix} 2\\ -1\\ -2\\ \end{pmatrix},$$ respectively. (It also tells you that the distance between the two lines is the norm of $\frac3{17}(-3\ {-2}\ {-2})$, or $\frac3{\sqrt{17}}$ as you indicated.)
Any point $$P$$ on line $$\vec{AB}$$ given $$u$$ can be represented as $$\begin{pmatrix} 1-2u \\ 2u\\ 2+u \end{pmatrix}$$ , whilst point any point $$Q$$ on line $$\vec{CD}$$ given $$v$$ can be represented as $$\begin{pmatrix} 2v \\ 1-v\\ 1-2v \end{pmatrix}$$, $$\vec{PQ}=$$ $$\begin{pmatrix} -1+2u+2v \\ 1-2u-v\\ -1-u-2v \end{pmatrix}$$
You can interpret the shortest distance between $$\vec{AB}$$ and $$\vec{CD}$$ at points $$P$$ and $$Q$$ as a separate line that is perpendicular to both $$\vec{AB},\vec{CD}$$ at the same time, similar to how the shortest distance from a point to a line functions: $$\vec{PQ} \perp \vec{AB} \\ \vec{PQ} \perp \vec{CD}$$ The dot product between two perpendicular vectors is 0, which gives: $$\vec{PQ} \cdot \vec{AB} = \begin{pmatrix} -1+2u+2v \\ 1-2u-v\\ -1-u-2v \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 2\\ 1 \end{pmatrix} = 0 \\ \vec{PQ} \cdot \vec{CD} = \begin{pmatrix} -1+2u+2v \\ 1-2u-v\\ -1-u-2v \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1\\ -2 \end{pmatrix} = 0$$ Which gives the following system of equations \begin{aligned} 3-9u-8v = 0\\ -1+8u+9v=0 \end{aligned} Solving for $$u,v$$ gives $$u=\frac{19}{17},v=-\frac{15}{17}$$, which can be substituted back to calculate the coordinates of $$P$$ and $$Q$$, giving $$P=(\frac{38}{17}, \frac{53}{17}, -\frac{21}{17}), Q=(-\frac{30}{17}, \frac{32}{17}, \frac{47}{17})$$