Combine parallelogram vectors, an internal angle and a nearest point

Follow the vertices O→A→B→C around the boundary, anticlockwise in the source schematic. Opposite sides OC and AB are equal and parallel in the same sense, so $\overrightarrow{AB}=\overrightarrow{OC}$. By the triangle law, $\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OA}+\overrightarrow{OC}=(3,11,11)+(2,-1,10)=(5,10,21)$. The Teacher labelling note helps identify adjacent vertices; the cyclic order is what matters, since the apparent clockwise direction changes with the viewpoint.

$\overrightarrow{AO}=(-3,-11,-11)$ and $\overrightarrow{AB}=(2,-1,10)$. Their dot product is $-6+11-110=-105$ and their magnitudes are $\sqrt{251}$ and $\sqrt{105}$. Thus $-105=\sqrt{251}\sqrt{105}\cos\angle OAB$, giving $\angle OAB=\cos^{-1}[-105/(\sqrt{251}\sqrt{105})]=130^\circ$ to the nearest degree. This is the internal angle, so retain the negative dot product rather than forcing an acute answer.

$\mathbf r=\overrightarrow{OA}+\lambda\overrightarrow{AB}=(3,11,11)+\lambda(2,-1,10)$, $\lambda\in\mathbb R$.

For D to lie on AB, $(-1,13,-9)=(3,11,11)+\lambda(2,-1,10)$. The three equations are $-1=3+2\lambda$, $13=11-\lambda$, $-9=11+10\lambda$. Each gives λ=−2. Therefore D lies on AB, as required.

Let F be the foot from O. Since F is on AB, $\overrightarrow{OF}=(3+2\lambda,11-\lambda,11+10\lambda)$. The nearest point satisfies $\overrightarrow{OF}\cdot(2,-1,10)=0$. Expanding gives $6+4\lambda-11+\lambda+110+100\lambda=0$, or $105+105\lambda=0$, so λ=−1. Therefore $\overrightarrow{OF}=(3+2(-1),11-(-1),11+10(-1))=(1,12,1)$. Its dot product with AB is $2-12+10=0$, confirming perpendicularity. The foot is before A on the infinite line, as in the source figure.