The area of a triangle whose vertices are $A(1, -1, 2)$,$B(2, 1, -1)$ and $C(3, -1, 2)$ is

If $A(1,-1,2)$,$B(5,7,-6)$,$C(3,4,-10)$,and $D(-1,-4,-2)$ are the vertices of a quadrilateral $ABCD$,then its area is:

The line passing through the point $i + 3j + 2k$ and perpendicular to the lines $r = (i + 2j - k) + \lambda (2i + j + k)$ and $r = (2i + 6j + k) + \mu (i + 2j + 3k)$ is:

Let $\bar{a}$,$\bar{b}$,and $\bar{c}$ be unit vectors. Suppose that $\bar{a} \cdot \bar{b} = \bar{a} \cdot \bar{c} = 0$ and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{6}$. Then $\bar{a}$ is equal to:

Let $\bar{u}=\hat{i}+\hat{j}$,$\bar{v}=\hat{i}-\hat{j}$ and $\bar{w}=\hat{i}+2\hat{j}+3\hat{k}$. If $\hat{n}$ is a unit vector such that $\bar{u} \cdot \hat{n}=0$ and $\bar{v} \cdot \hat{n}=0$,then $|\bar{w} \cdot \hat{n}|$ is equal to

If $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + \lambda \hat{j} + \mu \hat{k}) = \vec{0}$,then $\lambda$ and $\mu$ are respectively: