The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{k} \times \hat{i}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ is . . . . . . .

Two adjacent sides of a parallelogram $PQRS$ are given by $\vec{PQ} = \hat{i} + \hat{k}$ and $\vec{PS} = \hat{i} - \hat{j}$. If the side $PS$ is rotated about the point $P$ by an acute angle $\alpha$ in the plane of the parallelogram so that it becomes perpendicular to the side $PQ$,then $\sin^2(\frac{5\alpha}{2}) - \sin^2(\frac{\alpha}{2})$ is equal to:

If $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=3 \hat{i}+\hat{j}$ are vectors such that $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then the value of $\lambda$ is

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=\sqrt{14}$,$|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^2$ is equal to $...........$.

If $a = (1, -1, 2)$,$b = (-2, 3, 5)$,$c = (2, -2, 4)$ and $i$ is the unit vector in the $x$-direction,then $(a - 2b + 3c) \cdot i = $

Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c}+\vec{d}|=\sqrt{29}$ and $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$. If $\lambda_1, \lambda_2$ $(\lambda_1 > \lambda_2)$ are the possible values of $(\vec{c}+\vec{d}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k})$,then the equation $K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+(3K+\frac{\lambda_{2}}{2})y^{2}-8x+12y+\lambda_{2}=0$ represents a circle,for $K$ equal to: