有 $\overset{\large\rightharpoonup}{AB}=(1,0,-1)$、$\overset{\large\rightharpoonup}{AC}=(0,2,1)$，故 $\overset{\large\rightharpoonup}{AB}\times\overset{\large\rightharpoonup}{AC}=(2,-1,2)$，長度為 $3$。$\overset{\large\rightharpoonup}{AH}$ 中平行平面 $E$ 的部分為 $\dfrac{2}{3}\overset{\large\rightharpoonup}{AB}-\dfrac{1}{3}\overset{\large\rightharpoonup}{AC}$，法向量部分為 $3(\overset{\large\rightharpoonup}{AB}\times\overset{\large\rightharpoonup}{AC})$。四面體體積為 $\dfrac{|(\overset{\large\rightharpoonup}{AB}\times\overset{\large\rightharpoonup}{AC})\cdot\overset{\large\rightharpoonup}{AH}|}{6}=\dfrac{3\cdot9}{6}=\dfrac{9}{2}$。反射時平面內分量不變、法向量分量變號，所以 $\overset{\large\rightharpoonup}{AH'}=\dfrac{2}{3}\overset{\large\rightharpoonup}{AB}-\dfrac{1}{3}\overset{\large\rightharpoonup}{AC}-3(\overset{\large\rightharpoonup}{AB}\times\overset{\large\rightharpoonup}{AC})=\left(-\dfrac{16}{3},\dfrac{7}{3},-7\right)$，故 $H'=\left(-\dfrac{16}{3},\dfrac{4}{3},-8\right)$。投影點為 $A+\dfrac{2}{3}\overset{\large\rightharpoonup}{AB}-\dfrac{1}{3}\overset{\large\rightharpoonup}{AC}$，其相對於 $A,B,C$ 的仿射係數含負係數，故不在 $\triangle ABC$ 的內部。