Giải thích các bước giải:
Gọi $D(x,y,z)$
$\to\vec{AD}=(x+2,y,z),\vec{BD}=(x,y+2,z),\vec{CD}=(x,y,z+2)$
$\to\vec{AD}.\vec{BD}=\vec{BD}.\vec{CD}=\vec{CD}.\vec{AD}=0$
$\to\begin{cases}(x+2)x+y(y+2)+z^2=0\\x^2+y(y+2)+z(z+2)=0\\x(x+2)+y^2+z(z+2)=0\end{cases}$
$\to\begin{cases}(x+2)x+y(y+2)+z^2-(x^2+y(y+2)+z(z+2))=0\\x^2+y(y+2)+z(z+2)-(x(x+2)+y^2+z(z+2))=0\\x(x+2)+y^2+z(z+2)-((x+2)x+y(y+2)+z^2)=0\end{cases}$
$\to\begin{cases}2x-2z=0\\-2x+2y=0\\-2y+2z=0\end{cases}$
$\to x=y=z$
$\to x(x+2)+x(x+2)+x^2=0\to x=-\dfrac43$ vì $x\ne 0$
$\to D(-\dfrac43,-\dfrac43,-\dfrac43)$
