What is the area of the parallelogram whose sides are represented...

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$.

The area of the parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} \times \vec{b}|$.

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} \text{ and } \hat{j} \text{ and } \hat{k} \\ 1 \text{ and } 2 \text{ and } 3 \\ 2 \text{ and } 1 \text{ and } 2 \end{vmatrix}$

$= \hat{i}(4 - 3) - \hat{j}(2 - 6) + \hat{k}(1 - 4)$

$= \hat{i} + 4\hat{j} - 3\hat{k}$

The magnitude of the cross product is:

$|\vec{a} \times \vec{b}| = \sqrt{1^2 + 4^2 + (-3)^2}$

$= \sqrt{1 + 16 + 9} = \sqrt{26}$

The area of the parallelogram is $\sqrt{26} square units.<br><br>Answer: \sqrt{26}$ square units