If $A, B, C, D$ are the points with position vectors $\hat{i}-\hat{j}+\hat{k}, 2 \hat{i}-\hat{j}+3 \hat{k}, 2 \hat{i}-3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$ respectively,find the projection of $\overrightarrow{AB}$ on $\overrightarrow{CD}$.

(D) Given position vectors are $\overrightarrow{OA} = \hat{i}-\hat{j}+\hat{k}$,$\overrightarrow{OB} = 2\hat{i}-\hat{j}+3\hat{k}$,$\overrightarrow{OC} = 2\hat{i}-3\hat{k}$,and $\overrightarrow{OD} = 3\hat{i}-2\hat{j}+\hat{k}$.
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2-1)\hat{i} + (-1 - (-1))\hat{j} + (3-1)\hat{k} = \hat{i} + 0\hat{j} + 2\hat{k}$.
$\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = (3-2)\hat{i} + (-2-0)\hat{j} + (1 - (-3))\hat{k} = \hat{i} - 2\hat{j} + 4\hat{k}$.
The projection of $\overrightarrow{AB}$ on $\overrightarrow{CD}$ is given by $\frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{CD}|}$.
$\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(1) + (0)(-2) + (2)(4) = 1 + 0 + 8 = 9$.
$|\overrightarrow{CD}| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.
Therefore,the projection is $\frac{9}{\sqrt{21}} = \frac{9\sqrt{21}}{21} = \frac{3\sqrt{21}}{7}$ units.