Given $\overrightarrow{OA}=\hat i+\hat j+\hat k$, $\overrightarrow{OB}=2\hat i+\hat j+3\hat k$, $\overrightarrow{OC}=2\hat i-3\hat k$ and $\overrightarrow{OD}=3\hat i+2\hat j+\hat k$
Let us determine $ \overrightarrow{AB}$ and $ \overrightarrow{CD}$
$ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$\qquad=(2\hat i+\hat j+3\hat k)-(\hat i+\hat j+\hat k)$
$\qquad=\hat i+2\hat k$
$ \overrightarrow{CD}=\overrightarrow{OD}-\overrightarrow{OC}$
$\qquad=(3\hat i+2\hat j+\hat k)-(2\hat i-3\hat k)$
$\qquad=\hat i+2\hat j+ 4 \hat k$
Projection of $ \overrightarrow {AB} $ along $ \overrightarrow {CD}$ is
$\large\frac{\overrightarrow {AB}.\overrightarrow {CD}}{|\overrightarrow {CD}|} =\large\frac{(\hat i+2 \hat k).(\hat i+2 \hat j +4 \hat k)}{\sqrt {1^2+2^2+(-2)^2}}$
$=\large\frac{1+8}{\sqrt {1+4+4}}=\frac{9}{3}$
$=3$
Hence the projection of $ \overrightarrow{AB} \: along \: \overrightarrow{CD}$ is 3