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Q.
The area of the triangle formed by the adjacent sides with $\vec{A}=-3\hat{i}+2\hat{j}-4\hat{k}$ and $\vec{B}=-\hat{i}+2\hat{j}+\hat{k}$ is
Motion in a Plane
Solution:
$\vec{A}\times\vec{B}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ -3&2&-4\\ -1&2&1\end{vmatrix}$
$=\hat{i}\left[2+8\right]+\hat{j}\left[4+3\right]+\hat{k}\left[-6+2\right]$
$=10\hat{i}+7\hat{j}-4\hat{k}$
$\therefore \left|\vec{A}\times\vec{B}\right|=\sqrt{10^{2}+7+4^{2}}$
$=\sqrt{100+49+16}=\sqrt{165}$
Area of triangle $=\frac{1}{2}\left|\vec{A}\times\vec{B}\right|$
$=\frac{\sqrt{165}}{2}$ units