Step 1: Dot product of $\vec{A}$ and $\vec{B}$.
\[ \vec{A}\cdot \vec{B} = 9(11) + (-5)(4) + 2(1) = 99 - 20 + 2 = 81 \neq 0 \] So, $\vec{A}$ and $\vec{B}$ are not perpendicular. $\Rightarrow$ (C) false.
Step 2: Scalar triple product.
\[ \vec{A}\cdot(\vec{B}\times \vec{C}) \] \[ \vec{B}\times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 11 & 4 & 1 \\ -7 & 14 & -3 \end{vmatrix} = (-4-14)\hat{i} - (-33+7)\hat{j} + (154+28)\hat{k} \] \[ = -18\hat{i} + 26\hat{j} + 182\hat{k} \] Now, \[ \vec{A}\cdot(\vec{B}\times \vec{C}) = 9(-18) + (-5)(26) + 2(182) = -162 -130 + 364 = 72 \] Wait – not zero. Let's recheck carefully. \[ \vec{B}\times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 11 & 4 & 1 \\ -7 & 14 & -3 \end{vmatrix} \] \[ = (4\cdot -3 - 1\cdot 14)\hat{i} - (11\cdot -3 - 1\cdot -7)\hat{j} + (11\cdot 14 - 4\cdot -7)\hat{k} \] \[ = (-12 -14)\hat{i} - (-33+7)\hat{j} + (154+28)\hat{k} \] \[ = -26\hat{i} + 26\hat{j} + 182\hat{k} \] Now dot with $\vec{A}$: \[ \vec{A}\cdot(\vec{B}\times \vec{C}) = 9(-26) + (-5)(26) + 2(182) = -234 -130 + 364 = 0 \] So scalar triple product = 0 $\Rightarrow$ (A) and (B) true.
Step 3: Parallel check.
$\vec{A}\times \vec{B}$ must be checked against $\vec{C}$. \[ \vec{A}\times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 9 & -5 & 2 \\ 11 & 4 & 1 \end{vmatrix} = (-5\cdot 1 - 2\cdot 4)\hat{i} - (9\cdot 1 - 2\cdot 11)\hat{j} + (9\cdot 4 - (-5)\cdot 11)\hat{k} \] \[ = (-5 -8)\hat{i} - (9 -22)\hat{j} + (36+55)\hat{k} = -13\hat{i} + 13\hat{j} + 91\hat{k} \] Compare with $\vec{C} = -7\hat{i}+14\hat{j}-3\hat{k}$. Not scalar multiples. $\Rightarrow$ not parallel. So (D) false. \[ \boxed{\text{Correct statements: (A) and (B)}} \]
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