$A$ unit vector perpendicular to the vectors $\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $\vec{b} = 3 \hat{j} + 2 \hat{k}$ is

The area of the parallelogram whose adjacent sides are $\vec{a} = \hat{i} - \hat{k}$ and $\vec{b} = 2\hat{j} + 3\hat{k}$ is

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$,$\overrightarrow{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$,$\overrightarrow{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\overrightarrow{d}$ be a vector such that $\overrightarrow{b} \times \overrightarrow{d} = \overrightarrow{c} \times \overrightarrow{d}$ and $\overrightarrow{a} \cdot \overrightarrow{d} = 4$. Then $|(\overrightarrow{a} \times \overrightarrow{d})|^2$ is equal to . . . . . . .

Let a vector $\vec{a}$ have a magnitude $9$. Let a vector $\vec{b}$ be such that for every $(x, y) \in \mathbb{R} \times \mathbb{R} \setminus \{(0,0)\}$,the vector $(x \vec{a} + y \vec{b})$ is perpendicular to the vector $(6y \vec{a} - 18x \vec{b})$. Then the value of $|\vec{a} \times \vec{b}|$ is equal to:

$A$ unit vector which is perpendicular to $i + 2j - 2k$ and $-i + 2j + 2k$ is

If $\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=-3 \hat{i}+5 \hat{j}-4 \hat{k}$ and $\vec{c}=6 \hat{i}-4 \hat{j}+5 \hat{k}$,then $(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=$