The area of the parallelogram is determined by the magnitude of the cross product of its adjacent sides. The formula is given as:
\[
\text{Area} = |\mathbf{a} \times \mathbf{b}|
\]
First, we compute the cross product \( \mathbf{a} \times \mathbf{b} \). Given the vectors \( \mathbf{a} = 2 \hat{i} - \hat{j} + \hat{k} \) and \( \mathbf{b} = \hat{i} + 3 \hat{j} - \hat{k} \), the cross product can be expressed using a determinant:
\[
\mathbf{a} \times \mathbf{b} = \left|
\begin{matrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
1 & 3 & -1
\end{matrix}
\right|
\]
Expanding this determinant yields:
\[
\mathbf{a} \times \mathbf{b} = \hat{i} \left[ (-1)(-1) - (1)(3) \right] - \hat{j} \left[ (2)(-1) - (1)(1) \right] + \hat{k} \left[ (2)(3) - (-1)(1) \right]
\]
\[
= \hat{i} (1 - 3) - \hat{j} (-2 - 1) + \hat{k} (6 + 1)
= -2 \hat{i} + 3 \hat{j} + 7 \hat{k}
\]
Next, we calculate the magnitude of this cross product:
\[
|\mathbf{a} \times \mathbf{b}| = \sqrt{(-2)^2 + 3^2 + 7^2} = \sqrt{4 + 9 + 49} = \sqrt{62}
\]
Therefore, the area of the parallelogram is \( \sqrt{62} \).