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

If $A=(1,-1,2)$, $B=(3,4,-2)$, $C=(0,3,2)$ and $D=(3,5,6)$, then the angle between the lines $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is (in $^{\circ}$)

If $\overrightarrow{a} \cdot \hat{i} = \overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = \overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$,then $\overrightarrow{a}$ is equal to :

Find the distance between the lines $l_{1}$ and $l_{2}$ given by $\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$ and $\vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+6 \hat{k})$.

Evaluate the product $(3 \vec{a}-5 \vec{b}) \cdot (2 \vec{a}+7 \vec{b})$.

Let $\vec{a} = 2\hat{i} + \hat{k}$, $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$, then $\vec{r} \cdot \vec{c} = $