\( \left. \begin{array} { c } { x = \langle 3,1 , b \rangle \quad b = \langle - 1,2,3 \rangle \quad c = 2 i + b j - 2 k } \\ { a \cdot ( b \times c ) = a \times b + a \cdot c } \end{array} \right. \)

To solve the problem, we need to analyze the given vectors and the equation involving the dot and cross products. 1. **Identify the vectors**: - \( x = \langle 3, 1, b \rangle \) suggests that \( b \) is a scalar, but it is also defined as \( b = \langle -1, 2, 3 \rangle \). This means \( x = \langle 3, 1, -1 \rangle \). - The vector \( c \) is given as \( c = 2i + bj - 2k \). Substituting \( b = \langle -1, 2, 3 \rangle \) into this expression is not straightforward since \( b \) is a vector, not a scalar. We will treat \( b \) as a vector in the context of \( c \). 2. **Express \( c \)**: - We can express \( c \) in terms of its components: \[ c = 2i + (-1)j - 2k = \langle 2, -1, -2 \rangle \] 3. **Calculate the cross product \( b \times c \)**: - First, we need to compute \( b \times c \): \[ b = \langle -1, 2, 3 \rangle, \quad c = \langle 2, -1, -2 \rangle \] - The cross product \( b \times c \) can be calculated using the determinant of a matrix: \[ b \times c = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 3 \\ 2 & -1 & -2 \end{vmatrix} \] - Expanding this determinant: \[ = \mathbf{i} \begin{vmatrix} 2 & 3 \\ -1 & -2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -1 & 3 \\ 2 & -2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -1 & 2 \\ 2 & -1 \end{vmatrix} \] - Calculating the 2x2 determinants: \[ = \mathbf{i} (2 \cdot -2 - 3 \cdot -1) - \mathbf{j} (-1 \cdot -2 - 3 \cdot 2) + \mathbf{k} (-1 \cdot -1 - 2 \cdot 2) \] \[ = \mathbf{i} (-4 + 3) - \mathbf{j} (2 - 6) + \mathbf{k} (1 - 4) \] \[ = \mathbf{i} (-1) - \mathbf{j} (-4) + \mathbf{k} (-3) \] \[ = \langle -1, 4, -3 \rangle \] 4. **Calculate the dot product \( a \cdot (b \times c) \)**: - We need to find \( a \) to compute \( a \cdot (b \times c) \). However, \( a \) is not defined in the problem. Assuming \( a = x = \langle 3, 1, -1 \rangle \): \[ a \cdot (b \times c) = \langle 3, 1, -1 \rangle \cdot \langle -1, 4, -3 \rangle \] \[ = 3 \cdot (-1) + 1 \cdot 4 + (-1) \cdot (-3) \] \[ = -3 + 4 + 3 = 4 \] 5. **Calculate \( a \times b \)**: - Now we compute \( a \times b \): \[ a \times b = \langle 3, 1, -1 \rangle \times \langle -1, 2, 3 \rangle \] - Using the determinant: \[ = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & -1 \\ -1 & 2 & 3 \end{vm