Problem source

The position vectors of the points $A\,$, $B\,$ and $P$ with respect to an origin $O$ are $a{\bf i}\,$,  $b{\bf j}\,$ and  $l{\bf i}+m{\bf j}+n{\bf k}\,$, respectively, where $a$, $b$, and $n$ are all non-zero. The points $E$, $F$, $G$ and $H$ are the midpoints of $OA$, $BP$, $OB$ and  $AP$, respectively.  
Show that the lines $EF$ and $GH$  intersect.
Let $D$ be the point with position vector $d{\bf k}$, where $d$ is non-zero, and let $S$ be the point of intersection of $EF$ and $GH.$
The point $T$ is such that the mid-point of $DT$ is $S$. 
Find the position vector of $T$ and hence find $d$ in terms of $n$
if $T$ lies  in the plane $OAB$.

Solution source

$E = \langle \frac{a}{2}, 0,0 \rangle, F = \langle \frac{l}{2}, \frac{m+b}{2}, \frac{n}{2} \rangle, G = \langle 0, \frac{b}{2}, 0 \rangle, H = \langle \frac{a+l}{2}, \frac{m}{2}, \frac{n}{2} \rangle$

Note that the midpoint of $EF$ and $GH$ are both $\langle \frac{a+l}{4}, \frac{m+b}{4}, 
\frac{n}{4} \rangle$, so clearly they must intersect at this point.

The vector we just found is $S$, and $\mathbf{t} = \mathbf{d} + 2(\mathbf{s}-\mathbf{d}) = 2\mathbf{s} - \mathbf{d}$. Therefore $T = \langle \frac{a+l}{2}, \frac{m+b}{2}, \frac{n-2d}{2} \rangle$. If $T$ lies in the plane $OAB$ then $n - 2d = 0$ ie $d = \frac{n}{2}$