Step 1: Position vectors:
Each position vector uses the coordinates as components.
$\vec{a} = 3\hat{i} - \hat{j} + 2\hat{k}$, $\vec{b} = 6\hat{i} - 3\hat{j} + 6\hat{k}$.
$\vec{c} = 4\hat{i} - 3\hat{j} + \hat{k}$, $\vec{d} = \hat{i} - \hat{j} - 3\hat{k}$.
Step 2: Use the diagonals:
A quadrilateral is a parallelogram if its diagonals bisect each other.
Mid-point of AC: $\left( \frac{3 + 4}{2}, \frac{-1 - 3}{2}, \frac{2 + 1}{2} \right) = \left( \frac{7}{2}, -2, \frac{3}{2} \right)$.
Mid-point of BD: $\left( \frac{6 + 1}{2}, \frac{-3 - 1}{2}, \frac{6 - 3}{2} \right) = \left( \frac{7}{2}, -2, \frac{3}{2} \right)$.
The mid-points coincide, so ABCD is a parallelogram.
Step 3: Find the area with sides AB and BC:
Take sides $\vec{AB} = 3\hat{i} - 2\hat{j} + 4\hat{k}$ and $\vec{BC} = -2\hat{i} + 0\hat{j} - 5\hat{k}$.
Their cross product is \[ \vec{AB}\times\vec{BC} = 10\hat{i} + 7\hat{j} - 4\hat{k} \]
The area of the parallelogram equals the magnitude of this vector.
Step 4: Magnitude:
\[ \sqrt{10^2 + 7^2 + (-4)^2} = \sqrt{165} \]
The value is about 12.85 square units.
Final Answer:
Area of ABCD $= \sqrt{165}$ square units.
\[ \boxed{\sqrt{165}} \]