Multiplying by \( b \) in the given relation, we have: \[ b \cdot (a \times b) + b \cdot c = 0 \implies b \cdot c = 0 \] Also given that \( a \cdot b = 3 \), suppose: \[ b = x \, \mathbf{i} + y \, \mathbf{j} + z \, \mathbf{k} \] Thus: \[ y - z = 3 \quad \text{and} \quad x - y - z = 0 \] From these equations: \[ x = 2z + 3, \quad y = z + 3 \] So the vector: \[ b = (2z + 3) \, \mathbf{i} + (z + 3) \, \mathbf{j} + z \, \mathbf{k} \] Now, again from the relation \((a \times b) + c = 0\), we get: \[ z = -2 \] Thus, the vector: \[ b = -\mathbf{i} + \mathbf{j} - 2 \mathbf{k} \]