If $i-j+2k,2i+j-k$ and $3i-j+2k$ are position vectors of vertices of a triangle ,then its area is

The correct option is A $\sqrt{13}$
Given

$\overrightarrow{OA}=\hat{i}-\hat{j}+2\hat{k}$

$\overrightarrow{OB}=2\hat{i}+\hat{j}-\hat{k}$

$\overrightarrow{OC}=3\hat{i}-\hat{j}+2\hat{k}$

Now,

Area of triangle $=\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{BC}\right|$

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

$=\left(2-1\right)\hat{i}+\left(1+1\right)\hat{j}+\left(-1-2\right)\hat{k}$

$=\hat{i}+2\hat{j}-3\hat{k}$

And,

$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$

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

$=\hat{i}-2\hat{j}+3\hat{k}$

Now

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

$=\hat{i}\left(6-6\right)-\hat{j}\left(3+3\right)+\hat{k}\left(-2-2\right)$

$=0-6\hat{j}-4\hat{k}$

So,

$\left|\overrightarrow{AB}\times \overrightarrow{BC}\right|=\sqrt{{\left(6\right)}^2+{\left(4\right)}^2}=\sqrt{52}$

Hence, area of triangle is $\frac{1}{2}$$\sqrt{52}$=$\sqrt{13}$

Hence, the option $D$ is the correct answer.