#### 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: A. 9 square units B. 18 square units C. 27 square units D. 81 square units

Given, A (0, 4, 1), B (2, 3, -1), C (4, 5, 0) and D (2, 6, 2);

$\vec{AB}=\left ( 2-0 \right )\hat{i}+\left ( 3-4 \right )\hat{j}+\left ( -1-1 \right )\hat{k}=2\hat{i}-\hat{j}-2\hat{k}$

$\vec{BC}=\left ( 4-2\right )\hat{i}+\left ( 5-3 \right )\hat{j}+\left ( 0-(-1) \right )\hat{k}=2\hat{i}+2\hat{j}+\hat{k}$

$\vec{CD}=\left ( 2-4\right )\hat{i}+\left ( 6-5 \right )\hat{j}+\left ( 2-0 \right )\hat{k}=-2\hat{i}+\hat{j}+2\hat{k}=-\vec{AB}$

$\vec{DA}=\left ( 0-2\right )\hat{i}+\left ( 4-6 \right )\hat{j}+\left ( 1-2 \right )\hat{k}=-2\hat{i}-2\hat{j}-1\hat{k}=-\vec{BC}$

Since opposite vectors of this parallelogram are equal and opposite, ABCD is a parallelogram and we know the area of a parallelogram is | AB x CD|

$=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & -1 & -2 \\ 2 & 2 & 1 \end{vmatrix}$

$=\left | \hat{i}\left ( -1+4 \right )+\hat{j}\left ( -4-2 \right )+\hat{k}\left ( 4+2 \right ) \right |$

$=\left | 3\hat{i}-6\hat{j}+6\hat{k} \right |$

$=\sqrt{3^{2}+(-6)^{2}+6^{2}}=\sqrt{81}$

= 9 square units (Option A).