The area of the triangle whose vertices are A( 1,-1, 2) ,B(2,1,-1) and C(3,-1,2) is

Solution:

$Area \hspace{0.2cm}of \hspace{0.2cm}\bigtriangleup ABC=\frac{1}{2}\left | \vec{BA}\times \vec{BC} \right |\\ \\ \Rightarrow \hspace{1cm} \vec{BA}=-i-2j+3k\\ \\ \Rightarrow \hspace{1cm} \vec{BC}=i-2j+3k \\ \\ \Rightarrow \hspace{1cm} Area =\frac{1}{2}\begin{vmatrix} i &j &k \\ -1 & -2 &3 \\ 1 &-2 &3 \end{vmatrix}=\frac{1}{2}\left | 6i+4k \right |=\sqrt{13}$