Let $\overrightarrow{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\hat{i}-2 \hat{j}-3 \hat{k}$ be two vectors. If $A_1$ is the area of the quadrilateral having $\vec{a}, \vec{b}$ as its diagonals and $A_2$ is the area of the parallelogram having $\overrightarrow{a}, \overrightarrow{b}$ as its two adjacent sides,then $A_1 \cdot A_2=$

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that $\vec{a} \times \vec{b} = 4\vec{c}$,$\vec{b} \times \vec{c} = 9\vec{a}$,and $\vec{c} \times \vec{a} = \alpha\vec{b}$,where $\alpha > 0$. If $|\vec{a}| + |\vec{b}| + |\vec{c}| = 36$,then $\alpha$ is equal to:

If $\vec{a}_1$ is the component of vector $\vec{a}$ along the direction of vector $\vec{b}$,and $\vec{a}_2$ is the component of $\vec{a}$ perpendicular to $\vec{b}$,then $\vec{a}_1 \times \vec{a}_2 = \dots$

Let $F=2 \hat{i}+2 \hat{j}+5 \hat{k}$,$A=(1,2,5)$,$B=(-1,-2,-3)$ and $BA \times F=4 \hat{i}+6 \hat{j}+2 \lambda \hat{k}$,then $\lambda=$

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 . . . . . . .

If $|\overrightarrow{a}|=2, |\overrightarrow{b}|=3$ and $\overrightarrow{a}, \overrightarrow{b}$ are mutually perpendicular,then the area of the triangle whose vertices are $\overrightarrow{0}, \overrightarrow{a}+\overrightarrow{b}, \overrightarrow{a}-\overrightarrow{b}$ is