Q. \( 1 \quad \) Find \( \vec{a} \cdot(\vec{b} \times \vec{c}) \), if \( \vec{a}=2 \hat{i}+\hat{j}+3 \hat{k} \), \( \vec{b}=-\hat{i}+2 \hat{j}+\hat{k} \) and \( \vec{c}=3 \hat{i}+\hat{j}+2 \hat{k} \).

To find \( \vec{a} \cdot (\vec{b} \times \vec{c}) \), we first need to compute the cross product \( \vec{b} \times \vec{c} \). Given: \[ \vec{a} = 2 \hat{i} + \hat{j} + 3 \hat{k} \] \[ \vec{b} = -\hat{i} + 2 \hat{j} + \hat{k} \] \[ \vec{c} = 3 \hat{i} + \hat{j} + 2 \hat{k} \] The cross product \( \vec{b} \times \vec{c} \) can be calculated using the determinant of a matrix formed by the unit vectors and the components of \( \vec{b} \) and \( \vec{c} \): \[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 3 & 1 & 2 \end{vmatrix} \] Calculating this determinant, we expand it as follows: \[ \vec{b} \times \vec{c} = \hat{i} \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 1 \\ 3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} \] Calculating each of the 2x2 determinants: 1. For \( \hat{i} \): \[ \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = (2)(2) - (1)(1) = 4 - 1 = 3 \] 2. For \( -\hat{j} \): \[ \begin{vmatrix} -1 & 1 \\ 3 & 2 \end{vmatrix} = (-1)(2) - (1)(3) = -2 - 3 = -5 \quad \Rightarrow \quad -(-5) = 5 \] 3. For \( \hat{k} \): \[ \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} = (-1)(1) - (2)(3) = -1 - 6 = -7 \] Putting it all together, we have: \[ \vec{b} \times \vec{c} = 3 \hat{i} + 5 \hat{j} - 7 \hat{k} \] Now we compute \( \vec{a} \cdot (\vec{b} \times \vec{c}) \): \[ \vec{a} \cdot (\vec{b} \times \vec{c}) = (2 \hat{i} + \hat{j} + 3 \hat{k}) \cdot (3 \hat{i} + 5 \hat{j} - 7 \hat{k}) \] Calculating the dot product: \[ = 2 \cdot 3 + 1 \cdot 5 + 3 \cdot (-7) \] \[ = 6 + 5 - 21 \] \[ = 11 - 21 = -10 \] Thus, the final result is: \[ \vec{a} \cdot (\vec{b} \times \vec{c}) = -10 \]