Space is equipped with an orthonormal coordinate system $(\mathrm{O}; \vec{\imath}, \vec{\jmath}, \vec{k})$. We consider the three points $\mathrm{A}(3;0;0)$, $\mathrm{B}(0;2;0)$ and $\mathrm{C}(0;0;2)$.

The objective of this exercise is to demonstrate the following property:
``The square of the area of triangle ABC is equal to the sum of the squares of the areas of the three other faces of the tetrahedron OABC''.

\textbf{Part 1: Distance from point O to the plane (ABC)}

\begin{enumerate}
  \item Demonstrate that the vector $\vec{n}(2;3;3)$ is normal to the plane (ABC).
  \item Demonstrate that a Cartesian equation of the plane (ABC) is: $2x + 3y + 3z - 6 = 0$.
  \item Give a parametric representation of the line $d$ passing through O and with direction vector $\vec{n}$.
  \item We denote H the point of intersection of the line $d$ and the plane (ABC). Determine the coordinates of point H.
  \item Deduce that the distance from point O to the plane (ABC) is equal to $\dfrac{3\sqrt{22}}{11}$.
\end{enumerate}

\textbf{Part 2: Demonstration of the property}

\begin{enumerate}
  \item Demonstrate that the volume of the tetrahedron OABC is equal to 2.
  \item Deduce that the area of triangle ABC is equal to $\sqrt{22}$.
  \item Demonstrate that for the tetrahedron OABC, ``the square of the area of triangle ABC is equal to the sum of the squares of the areas of the three other faces of the tetrahedron''. Recall that the volume of a tetrahedron is given by $V = \dfrac{1}{3}B \times h$ where $B$ is the area of a base of the tetrahedron and $h$ is the height relative to this base.
\end{enumerate}