Solve perpendicularity conditions and compare two area methods

The section formula gives

$\overrightarrow{OP}=\lambda\overrightarrow{OB}+(1-\lambda)\overrightarrow{OA}=\begin{pmatrix}1\\5\lambda-2\\5-5\lambda\end{pmatrix},$

and

$\overrightarrow{OQ}=\mu\overrightarrow{OD}+(1-\mu)\overrightarrow{OC}=\begin{pmatrix}10-12\mu\\3\mu+1\\3\mu+2\end{pmatrix}.$

Subtract the position vectors:

$\overrightarrow{PQ}=\begin{pmatrix}9-12\mu\\3-5\lambda+3\mu\\-3+5\lambda+3\mu\end{pmatrix}.$

Since $\overrightarrow{AB}=(0,5,-5)$, the equation $PQ\cdot AB=0$ is

$\begin{aligned}&5(3-5\lambda+3\mu)-5(-3+5\lambda+3\mu)\\ & =0.\end{aligned}$

This gives $\lambda=3/5$.

$\overrightarrow{CD}=(-12,3,3)$. Hence $PQ\cdot CD=0$ gives

$\begin{aligned}&-12(9-12\mu)+3(3-5\lambda+3\mu)\\ &{}+3(-3+5\lambda+3\mu)\\ & =0.\end{aligned}$

It simplifies to $-108+162\mu=0$, so $\mu=2/3$. Substitution gives $\overrightarrow{PQ}=(1,2,2)$, as required. In particular $P=(1,1,2)$ and $Q=(2,3,4)$.

$\overrightarrow{AQ}=Q-A=(1,5,-1)$ and $\overrightarrow{AB}=(0,5,-5)$. Therefore

$\overrightarrow{AB}\times\overrightarrow{AQ}=5\begin{pmatrix}4\\-1\\-1\end{pmatrix},$

and

$\begin{aligned}&\text{area}(ABQ)\\ & =\frac 12|AB\times AQ|\\ & =\frac 52\sqrt{18}\\ & =\frac{15\sqrt 2}{2}.\end{aligned}$

P lies on AB and PQ is perpendicular to AB, so PQ is the altitude from Q. The lengths are $|AB|=\sqrt{50}$ and $|PQ|=\sqrt 9=3$. Thus

$\begin{aligned}&\text{area}(ABQ)\\ & =\frac 12\sqrt{50}(3)\\ & =\frac{15\sqrt 2}{2}.\end{aligned}$

Both source triangle figures are represented in the two comparison panels.