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

In quadrilateral \(A B C D\), given that \(A(0,4,1), B(2,3,-1), C(4,5,0)\) and \(D(2,6,2)\).
\(\begin{aligned}
&\mathbf{A B}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\
&\mathbf{A D}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
&\mathbf{C B}=-2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\
&\mathbf{C D}=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}
\end{aligned}\)



Required area of quadrilateral
\(\begin{aligned}
&=\frac{1}{2}|\mathbf{A B} \times \mathbf{A D}|+\frac{1}{2}|\mathbf{C B} \times \mathbf{C D}| \\
&=\frac{1}{2}|3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}|+\frac{1}{2}|-3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}| \\
&=\frac{1}{2} \sqrt{3^{2}+(-6)^{2}+6^{2}}+\frac{1}{2} \sqrt{(-3)^{2}+6^{2}+(-6)^{2}} \\
&=\frac{1}{2} \times 9+\frac{1}{2} \times 9=9 \text { sq units }
\end{aligned}\)