Find the orthocenter

An equation for the plane through $A$ and perpendicular to $\overrightarrow{BC}=(-2,1,4)$ is $-2x+y+4z =12$ , while an equation for the plane through $B$ and perpendicular to $\overrightarrow{AC}=(2,2,2)$ is $x+y+z =9$ . As $\overrightarrow{AB} \times \overrightarrow{AC}=(6,-12,6)$ , the vector $(1,-2,1)$ is perpendicular to the triangle plane, hence an equation for this plane is $x-2y+z=0$ . The linear system consisting of the three equations above yields the coordinates of the orthocenter. The (unique) solution is $H=(5/2,3,7/2)$ and then two times the squared distance of the orthocenter from the origin is $55$ .