Answer
$\frac{\sqrt {21}}{2}$
Work Step by Step
We have $\vec{AB}= -\hat{i}+2\hat{j}$
and $\vec{AC}=\hat{j}-2\hat{k}$.
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}$.
Therefore, $|\vec{AB}\times\vec{AC}|=\sqrt {16+4+1}= \sqrt {21}$.
The area of the given triangle= $\frac{1}{2}|\vec{AB}\times\vec{AC}|=\frac{\sqrt {21}}{2}$
