If $A, B, C, D$ are four points in space,then $|\overline{AB} \times \overline{CD} + \overline{BC} \times \overline{AD} + \overline{CA} \times \overline{BD}| = \lambda \times (\text{Area of } \Delta ABC)$. Find $\lambda$.

If $a=\hat{i}+\hat{j}+\hat{k}$, $c=\hat{j}-\hat{k}$, $a \times b=c$ and $a \cdot b=3$, then $b$ is equal to

Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonals of a parallelogram having area $2 \sqrt{2}$. Let the angle between $\vec{a}$ and $\vec{b}$ be acute. Given $|\vec{a}|=1$ and $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$. If $\vec{c}=2 \sqrt{2}(\vec{a} \times \vec{b})-2 \vec{b}$,then find the angle between $\vec{b}$ and $\vec{c}$.

The unit vectors perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ are:

The points $A(a), B(b), C(c)$ will be collinear if

If the vertices of $\Delta ABC$ are $A=(2,3,5), B=(-1,3,2), C=(3,5,-2)$,then the area of the $\Delta ABC$ (in sq. units) is