If $\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{a}=2 \hat{i}+2 \hat{j}+p \hat{k}$, $|\vec{b}|=7$, $\vec{a} \cdot \vec{b}=4$ and $|\vec{a} \times \vec{b}|=5 \sqrt{17}$, then $p=$

If $a(\vec{\alpha} \times \vec{\beta}) + b(\vec{\beta} \times \vec{\gamma}) + c(\vec{\gamma} \times \vec{\alpha}) = \overrightarrow{0}$, where $a, b, c$ are non-zero scalars, then the vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are

The projection of vector $\vec{a}$ along vector $\vec{b}$ is:

If $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$,$|\vec{b}| = 5$,and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then the area of the triangle formed by these two vectors as two sides is

If $P=(0,1,2)$, $Q=(4,-2,1)$, and $O=(0,0,0)$, then $\angle POQ$ is equal to

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 . . . . . . .