The area of the parallelogram determined by the vectors $\hat{i}+2 \hat{j}+3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$ is

$8 \sqrt{3}$

$\vec{v_1}=\hat{i}+2 \hat{j}+3 \hat{i}$

$\vec{v}_2=3 \hat{i}-2 \hat{j}+\hat{k}$

finding $\vec{v_1} × \vec{v_2}$ 

using determinant

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

$\vec{v}_1 \times \vec{v}_2=8 \hat{i}+8 \hat{j}+(-8) \hat{k}$

so area of perallelogram $=\left|\vec{v}_1 \times \vec{v}_2\right|=\sqrt{8^2+8^2+8^2}$

$= 8 \sqrt{3}$