If $a \cdot i = a \cdot (i + j) = a \cdot (i + j + k)$, then $a = $

If the position vectors of the points $A$ and $B$ are $2\,i + 3\,j - k$ and $-2\,i + 3\,j + 4\,k$,then the line $AB$ is parallel to

If $\overline{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$,$\overline{b}=\hat{i}-2 \hat{j}+\hat{k}$,and $\overline{c}=\hat{i}+\hat{j}-\hat{k}$ are three vectors,and there exists a vector $\overline{r}$ such that $\overline{r} \times \overline{a}=\overline{b}$ and $\overline{r} \cdot \overline{c}=3$,then the value of $|\overline{r}|$ is:

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $|\overline{a}+\overline{b}+\overline{c}|=1$,$\overline{c}=\lambda(\overline{a} \times \overline{b})$ and $|\overline{a}|=\frac{1}{\sqrt{3}}, |\overline{b}|=\frac{1}{\sqrt{2}}, |\overline{c}|=\frac{1}{\sqrt{6}}$,then the angle between $\bar{a}$ and $\bar{b}$ is

If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$, $\vec{b} = \alpha\hat{i} + \beta\hat{j} + 2\hat{k}$ and $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$, then $\alpha + \beta$ is equal to

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}$,$\vec{a} \cdot \vec{b}=3$ and $|\vec{a} \times \vec{b}|^{2}=75$. Then $|\vec{a}|^{2}$ is equal to: