The area of a triangle having the points $A(1, 1,1), B(1, 2,3)$ and $C(2, 3 , 1)$ as its vertices is :

$\frac{1}{2}\sqrt{21}$

The correct answer is Option (4) → $\frac{1}{2}\sqrt{21}$

$\vec{AB}=\hat j+2\hat k$

$\vec{AC}=\hat i+2\hat j$

$\vec{AB}×\vec{AC}=\begin{vmatrix}\hat i&\hat j&\hat k\\0&1&2\\1&2&0\end{vmatrix}$

$-4\hat i+2\hat j-\hat k$

so $|\vec{AB}×\vec{AC}|=\sqrt{(-4)^2+2^2+(-1)^2}=\sqrt{21}$

Area (ΔABC) = $\frac{1}{2}|\vec{AB}×\vec{AC}|=\frac{\sqrt{21}}{2}$