If $|a \times b| = 4$ and $|a \cdot b| = 2$,then $|a|^2 |b|^2 = $

The angle between two consecutive sides $\vec{a}$ and $\vec{b}$ of a parallelogram is $\frac{\pi}{6}$. Given $\vec{a} = (2, -2, 1)$ and $\vec{b} = 2|\vec{a}|$,the area of the parallelogram is . . . . . . sq. units.

If the vectors $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$,and $\vec{c} = \lambda\hat{i} + \hat{j} + \mu\hat{k}$ are mutually perpendicular,then $(\lambda, \mu) = .......$

If $|a| = |b|$,then $(a + b) \cdot (a - b)$ is:

Let the angle $\theta, 0 < \theta < \frac{\pi}{2}$ between two unit vectors $\hat{a}$ and $\hat{b}$ be $\sin^{-1}\left(\frac{\sqrt{65}}{9}\right)$. If the vector $\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$,then the value of $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$ is

Show that the points $A (2 \hat{i}-\hat{j}+\hat{k})$,$B (\hat{i}-3 \hat{j}-5 \hat{k})$,and $C (3 \hat{i}-4 \hat{j}-4 \hat{k})$ are the vertices of a right-angled triangle.