The area of the quadrilateral ABCD, where A(0, 4, 1), B(2, 3, -1), C(4, 5, 0) and D(2, 6, 2) is equal to

The correct option is 9 sq. units.

To find the area of the quadrilateral with vertices $A(0,4,1)$, $B(2,3,-1)$, $C(4,5,0)$, and $D(2,6,2)$, we can use vector algebra.

Step 1: Find the adjacent side vectors
Let us calculate the vectors representing two adjacent sides of the quadrilateral, namely $\overrightarrow{AB}$ and $\overrightarrow{AD}$:
For side $\overrightarrow{AB}$:
$\overrightarrow{AB}=(x_B-x_A)\hat{i}+(y_B-y_A)\hat{j}+(z_B-z_A)\hat{k}$
$\overrightarrow{AB}=(2-0)\hat{i}+(3-4)\hat{j}+(-1-1)\hat{k}=2\hat{i}-\hat{j}-2\hat{k}$

For side $\overrightarrow{AD}$:
$\overrightarrow{AD}=(x_D-x_A)\hat{i}+(y_D-y_A)\hat{j}+(z_D-z_A)\hat{k}$
$\overrightarrow{AD}=(2-0)\hat{i}+(6-4)\hat{j}+(2-1)\hat{k}=2\hat{i}+2\hat{j}+\hat{k}$

Step 2: Identify the type of quadrilateral
Let us find the vectors for the other two sides $\overrightarrow{BC}$ and $\overrightarrow{DC}$:
$\overrightarrow{BC}=(4-2)\hat{i}+(5-3)\hat{j}+(0-(-1\left)\right)\hat{k}=2\hat{i}+2\hat{j}+\hat{k}$
$\overrightarrow{DC}=(4-2)\hat{i}+(5-6)\hat{j}+(0-2)\hat{k}=2\hat{i}-\hat{j}-2\hat{k}$

Since opposite sides are parallel and equal in length ($\overrightarrow{AB}=\overrightarrow{DC}$ and $\overrightarrow{AD}=\overrightarrow{BC}$), the quadrilateral ABCD is a parallelogram.

Step 3: Calculate the area of the parallelogram
The area of a parallelogram is equal to the magnitude of the cross product of its two adjacent side vectors:
$\text{Area}=|\overrightarrow{AB}\times \overrightarrow{AD}|$

First, we calculate the cross product $\overrightarrow{AB}\times \overrightarrow{AD}$ using the determinant of a 3x3 matrix:
$\overrightarrow{AB}\times \overrightarrow{AD}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& -2\\ 2& 2& 1\end{array}\right|$

Expanding the determinant:
$\overrightarrow{AB}\times \overrightarrow{AD}=\hat{i}\left[\right(-1\left)\right(1)-(-2\left)\right(2\left)\right]-\hat{j}\left[\right(2\left)\right(1)-(-2\left)\right(2\left)\right]+\hat{k}\left[\right(2\left)\right(2)-(-1\left)\right(2\left)\right]$
$\overrightarrow{AB}\times \overrightarrow{AD}=\hat{i}[-1+4]-\hat{j}[2+4]+\hat{k}[4+2]$
$\overrightarrow{AB}\times \overrightarrow{AD}=3\hat{i}-6\hat{j}+6\hat{k}$

Next, we find the magnitude of this cross product:
$|\overrightarrow{AB}\times \overrightarrow{AD}|=\sqrt{3^2+{(-6)}^2+6^2}$
$|\overrightarrow{AB}\times \overrightarrow{AD}|=\sqrt{9+36+36}$
$|\overrightarrow{AB}\times \overrightarrow{AD}|=\sqrt{81}=9$

Thus, the area of the quadrilateral ABCD is 9 sq. units.