1. **State the problem:** Find the cross product $\vec{a} \times \vec{b}$ where $\vec{a} = 14\hat{i} + 13\hat{j} + 15\hat{k}$ and $\vec{b} = -9\hat{i} + 15\hat{j} + 11\hat{k}$.\n\n2. **Formula for cross product:** The cross product of two vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ is given by the determinant:\n$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1)$$\n\n3. **Substitute values:**\n$a_1=14$, $a_2=13$, $a_3=15$, $b_1=-9$, $b_2=15$, $b_3=11$.\n\n4. **Calculate each component:**\n- $\hat{i}$ component: $13 \times 11 - 15 \times 15 = 143 - 225 = -82$\n- $\hat{j}$ component: $14 \times 11 - 15 \times (-9) = 154 + 135 = 289$\n- $\hat{k}$ component: $14 \times 15 - 13 \times (-9) = 210 + 117 = 327$\n\n5. **Apply signs:**\n$$\vec{a} \times \vec{b} = -82\hat{i} - 289\hat{j} + 327\hat{k}$$\n\n6. **Final answer:**\n$$\boxed{\vec{a} \times \vec{b} = -82\hat{i} - 289\hat{j} + 327\hat{k}}$$