Step-by-step Solution:

Let \[ b = x \hat{i} + y \hat{j} + z \hat{k}, \quad a \cdot b = 14. \] This implies: \[ 2x + 2y + z = 14 \quad \text{(Equation 1)}. \] Only the first choice satisfies this condition. Next, we can solve the question in detail using \( a \times b = 3 \hat{i} - \hat{j} - 8 \hat{k} \). We compute the cross product: \[ a \times b = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ x & y & z \end{array} \right| = (2z - y) \hat{i} - (2z - x) \hat{j} + (2z - 2x) \hat{k}. \] Comparing this with the given product \( a \times b = 3 \hat{i} - \hat{j} - 8 \hat{k} \), we have: \[ 2z - y = 3, \quad 2z - x = -1, \quad 2z - 2x = -8. \] Using equation (1), we find: \[ x = 5, \quad y = 1, \quad z = 2. \]