## Thomas' Calculus 13th Edition

$\dfrac{\sqrt {21}}{2}$
The area of the given triangle is $Area = \dfrac{1}{2}|\vec{AB}\times\vec{AC}|$ $$\vec{AB}= -\hat{i}+2\hat{j} \\ \vec{AC}=\hat{j}-2\hat{k}$$ Now, $$\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&2&0\\0&1&-2\end{vmatrix} \\ =-4\hat{i}-2\hat{j}-\hat{k}$$ and $$|\vec{AB}\times\vec{AC}|=\sqrt {16+4+1}= \sqrt {21}$$ Now, $Area= \dfrac{\sqrt {21}}{2}$