Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{3}$. If $\lambda \overrightarrow{a} + 2 \overrightarrow{b}$ and $3 \overrightarrow{a} - \lambda \overrightarrow{b}$ are perpendicular to each other,then the number of values of $\lambda$ in $[-1, 3]$ is:

Let a vector $\overrightarrow{a}=\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}$, $\lambda>0$, make an obtuse angle with the vector $\overrightarrow{b}=-\lambda^{2}\hat{i}+4\sqrt{2}\hat{j}+4\sqrt{2}\hat{k}$ and an angle $\theta$, $\frac{\pi}{6} < \theta < \frac{\pi}{2}$, with the positive $z$-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta)-\{\gamma\}$, then $\alpha+\beta+\gamma$ is equal to . . . . . . .

Let $\vec{b}$ and $\vec{c}$ be non-collinear vectors satisfying $\vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} \cdot \vec{b})\vec{b} = (4 - 2x - \sin y)\vec{b} + (x^2 - 1)\vec{c}$ and $(\vec{c} \cdot \vec{c})\vec{a} = \vec{c}$,then $x$ is equal to

If $a, b, c$ are distinct real numbers and $P, Q, R$ are three points whose position vectors are respectively $a \hat{i}+b \hat{j}+c \hat{k}$, $b \hat{i}+c \hat{j}+a \hat{k}$ and $c \hat{i}+a \hat{j}+b \hat{k}$, then $\angle Q P R=$

Find the angle between the vectors $\hat{i}-2 \hat{j}+3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$.

If $\vec{\alpha} = 3\hat{i} - \hat{k}$, $|\vec{\beta}| = \sqrt{5}$, and $\vec{\alpha} \cdot \vec{\beta} = 3$, then the area of the parallelogram for which $\vec{\alpha}$ and $\vec{\beta}$ are adjacent sides is: