If $a, b, c$ are position vectors of vertices of a triangle $ABC$,then the unit vector perpendicular to its plane is:

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

Given $a = i + j - k$,$b = -i + 2j + k$,and $c = -i + 2j - k$. $A$ unit vector perpendicular to both $a + b$ and $b + c$ is

If $\alpha$ is the angle between two vectors $p = 3\hat{i} + 4\hat{j} - \hat{k}$ and $q = 2\hat{i} - \hat{j} + \hat{k}$,then $\sin(\alpha) = $

If the position vectors of three points $A, B$ and $C$ are respectively $i + j + k, 2i + 3j - 4k$ and $7i + 4j + 9k$,then the unit vector perpendicular to the plane containing the triangle $ABC$ is

Let $\overline{a}=\hat{i}+2 \hat{j}-\hat{k}$ and $\overline{b}=\hat{i}+\hat{j}-\hat{k}$ be two vectors. If $\overline{c}$ is a vector such that $\overline{b} \times \overline{c}=\overline{b} \times \overline{a}$ and $\overline{c} \cdot \overline{a}=0$,then $\overline{c} \cdot \overline{b}$ is