$(a - b) \times (a + b) = $

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:

Let $\vec{p}=2 \hat{i}+3 \hat{j}+\hat{k}$ and $\vec{q}=\hat{i}+2 \hat{j}+\hat{k}$ be two vectors. If a vector $\vec{r}=(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k})$ is perpendicular to each of the vectors $(\vec{p}+\vec{q})$ and $(\vec{p}-\vec{q})$,and $|\vec{r}|=\sqrt{3}$,then $|\alpha|+|\beta|+|\gamma|$ is equal to $.....$

For any two vectors $a$ and $b$,if $a \times b = 0$,then

Let $A=(\alpha, 1, 2\alpha)$,$B=(3, 1, 2)$ and $C=4\hat{i}-\hat{j}+3\hat{k}$. If $AB \times C = 6\hat{i}+9\hat{j}-5\hat{k}$,then $\alpha^2+\alpha+5=$

The area of the parallelogram,whose diagonals are $\vec{d}_1 = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{d}_2 = \hat{i} + 3 \hat{j} - \hat{k}$,is equal to