Find the area of parallelogram and its missing vertex Given three radius-vectors: $OA(5; 1; 4), OB(6;2;3), OC(4;2;4)$, find the missing vertex $D$ and calculate the area of obtained parallelogram. 
My attempt:
Firstly, we are to find the vectors which form the parallelogram. $$AB = OA - OB = (-1; -1; 1)$$
$$AC=OA-OC=(1; -1; 0)$$
Since the parallelogram is spanned by $AB, AC$, cross product and the norm of resulting vector should be then calculated.
\begin{align*}
  &AB \times AC=
  {\begin{vmatrix}
    \vec{i} & \vec{j} & k \\ 
    -1 & -1 & 1 \\ 
    1 & -1 & 0 
  \end{vmatrix}} = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}
\end{align*}
$D$ is equidistant from $E$ and $B$, where $E$ is the point of diagonal's intersection. 
$$x_E = \frac{x_A+x_C}{2} = 9/2$$
$$y_E = \frac{y_A+y_C}{2} = 3/2$$
$$z_E = \frac{z_A+z_C}{2} = 4$$
\begin{cases} x_B - x_E = x_E - x_D\\ y_B - y_E = y_E - y_D \\ z_B - z_E = z_E - z_D \end{cases}
After plugging and solving, the final answer would be $\sqrt{6}$ and $D(3;1;5).$

A simpler way to obtain $D$ is $D = C+AB$.
