The vector $\vec{a} = (\alpha, 2, \beta)$ lies in the plane of the vectors $\vec{b} = (1, 1, 0)$ and $\vec{c} = (0, 1, 1)$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives the possible values of $\alpha$ and $\beta$?

Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be four vectors such that $\vec{a}$ is perpendicular only to $\vec{c}$. If the vector $\vec{b}$ is parallel to $(\vec{c}-\vec{d})$, then $\vec{c}$ is equal to:

If $|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 5$ such that each vector is perpendicular to the sum of the other two, then $|\vec{a} + \vec{b} + \vec{c}|$ is equal to

Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} - 2\hat{k}$ be three vectors. $A$ vector of the type $\vec{b} + \lambda \vec{c}$ for some scalar $\lambda$,whose projection on $\vec{a}$ is of magnitude $\sqrt{\frac{2}{3}}$ is

Let $\overrightarrow{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{k}$,and $\overrightarrow{c} = \hat{i} + 2\hat{j} - 3\hat{k}$ be three given vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$ and $\overrightarrow{r} \cdot \overrightarrow{b} = 0$,then $|\overrightarrow{r}|$ is equal to:

If the angle between the vectors $\vec{a} = 2\lambda^2 \hat{i} + 4\lambda \hat{j} + \hat{k}$ and $\vec{b} = 7\hat{i} - 2\hat{j} + \lambda \hat{k}$ is obtuse,then the values of $\lambda$ lie in: