The volume of the tetrahedron formed by the points $(1,1,1),(2,1,3),(3,2,2)$ and $(3,3,4)$ in cubic units is

The correct option is D $\frac{5}{6}$
Solution:

Let $A(1,1,1),B(2,1,3),C(3,2,2)$ and $D(3,3,4)$

$\overrightarrow{AB}=(2-1)\hat{i}+(1-1)\hat{j}+(3-1)\hat{k}=\hat{i}+2\hat{k}$

$\overrightarrow{AC}=(3-1)\hat{i}+(2-1)\hat{j}+(2-1)\hat{k}=2\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{AD}=(3-1)\hat{i}+(3-1)\hat{j}+(4-1)\hat{k}=2\hat{i}+2\hat{j}+3\hat{k}$

Volume of tetrahedron$=\frac{1}{6}\left|\begin{array}{ccc}1& 0& 2\\ 2& 1& 1\\ 2& 2& 3\end{array}\right|$ cubic units

$=\frac{1}{6}\left[1\right(3-2)+0(6-2)+2(4-2\left)\right]$ cubic units

$=\frac{1}{6}(1+0+4)$ cubic units

$=\frac{5}{6}$ cubic units

Hence, A is the correct option.