Find \( m \) for Volume of Parallelepiped

The volume of the parallelepiped is given by: \[ \text{Volume} = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| \]

Define the vectors: \[ \mathbf{a} = m\mathbf{i} + \mathbf{j} + \mathbf{k}, \quad \mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}, \quad \mathbf{c} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}. \]

Compute the cross product \( \mathbf{b} \times \mathbf{c} \): \[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix}. \]

Expanding the determinant: \[ \mathbf{b} \times \mathbf{c} = \mathbf{i}((-1)(-1) - (2)(1)) - \mathbf{j}((1)(-1) - (1)(1)) + \mathbf{k}((1)(2) - (1)(-1)). \] Simplifying: \[ \mathbf{b} \times \mathbf{c} = -\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}. \]

Compute the dot product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \): \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (m\mathbf{i} + \mathbf{j} + \mathbf{k}) \cdot (-\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}). \] Simplifying: \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = -m + 5. \]

Solve for \( m \) using the given volume of 4 cubic units: \[ | -m + 5 | = 4. \] This gives two cases: \[ 1. -m + 5 = 4 \implies m = 1, \] \[ 2. -m + 5 = -4 \implies m = -9. \]

Thus, the values of \( m \) are \( 1 \) and \( -9 \).