The area of the triangle formed by the adjacent sides with $\overrightarrow{A}=-3\hat{i}+2\hat{j}-4\hat{k} \mathrm{and} \overrightarrow{B}=-\hat{i}+2\hat{j}+\hat{k}$ is 

$\overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -3& 2& -4\\ -1& 2& 1\end{array}\right|$

       $=\hat{i}[2+8]-\hat{j}[-4-3]+\hat{k}[-6+2]=10\hat{i}+7\hat{j}-4\hat{k}$

$\therefore |\overrightarrow{A}\times \overrightarrow{B}|=\sqrt{{10}^2+7^2+4^2}=\sqrt{100+49+16}=\sqrt{165}$

Area of triangle $=\frac{1}{2}|\overrightarrow{A}\times \overrightarrow{B}|=\frac{\sqrt{165}}{2}\text{ units }$