The acute angle $\theta$ between the vector $\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k}$ and the plane containing the vectors $\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + 2\hat{k}$ is:

If $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + \lambda \hat{j} + \mu \hat{k}) = 0$,then $\lambda + \mu =$ . . . . . . .

Using vectors,find the area of the triangle $ABC$ with vertices $A(1, 2, 3)$,$B(2, -1, 4)$ and $C(4, 5, -1)$.

$a, b, c, d$ are coplanar vectors, then $(a \times b) \times (c \times d)$ is equal to

If either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0},$ then $\vec{a} \times \vec{b}=\overrightarrow{0} .$ Is the converse true? Justify your answer with an example.

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. If $\overline{c}$ is a vector such that $\overline{a} \cdot \overline{c}=|\overline{c}|$,$|\overline{c}-\overline{a}|=2 \sqrt{2}$ and the angle between $(\overline{a} \times \overline{b})$ and $\overline{c}$ is $\frac{\pi}{6}$,then $|(\overline{a} \times \overline{b}) \times \overline{c}|$ is