To determine the scalar triple product \( \vec{a} \cdot (\vec{b} \times \vec{c}) \), the initial step involves computing the cross product \( \vec{b} \times \vec{c} \). The provided vectors are: \( \vec{a} = \hat{i} + 2\hat{j} - \hat{k} \), \( \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k} \), and \( \vec{c} = -\hat{i} + 3\hat{j} + 2\hat{k} \).
Step 1: Calculate \( \vec{b} \times \vec{c} \)
The cross product is calculated using a determinant:
\[\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ -1 & 3 & 2 \end{vmatrix}\]
Expanding the determinant yields:
\[\vec{b} \times \vec{c} = \hat{i}((-1) \cdot 2 - 3 \cdot 3) - \hat{j}(2 \cdot 2 - 3 \cdot (-1)) + \hat{k}(2 \cdot 3 - (-1) \cdot (-1))\]
\[= \hat{i}(-2 - 9) - \hat{j}(4 + 3) + \hat{k}(6 - 1)\]
\[= -11\hat{i} - 7\hat{j} + 5\hat{k}\]
Step 2: Calculate \( \vec{a} \cdot (\vec{b} \times \vec{c}) \)
Next, the dot product with \( \vec{a} \) is performed:
\[\vec{a} \cdot (\vec{b} \times \vec{c}) = (\hat{i} + 2\hat{j} - \hat{k}) \cdot (-11\hat{i} - 7\hat{j} + 5\hat{k})\]
The calculation of each component is as follows:
\[= \hat{i} \cdot (-11\hat{i}) + 2\hat{j} \cdot (-7\hat{j}) + (-\hat{k}) \cdot (5\hat{k})\]
\[= -11 + (-14) - 5\]
\[= -30\]
Consequently, the value of \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) is \(-30\).