Find the unit vector which is perpendicular to the vector $5i + 2j + 6k$ and coplanar with the vectors $2i + j + k$ and $i - j + k$.

Let $\overrightarrow{a} = \alpha \hat{i} + 3 \hat{j} - \hat{k}$,$\overrightarrow{b} = 3 \hat{i} - \beta \hat{j} + 4 \hat{k}$ and $\overrightarrow{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ where $\alpha, \beta \in \mathbb{R}$,be three vectors. If the projection of $\overrightarrow{a}$ on $\overrightarrow{c}$ is $\frac{10}{3}$ and $\overrightarrow{b} \times \overrightarrow{c} = -6 \hat{i} + 10 \hat{j} + 7 \hat{k}$,then the value of $\alpha + \beta$ is equal to:

The unit vector perpendicular to the vector $\hat{i}-2 \hat{j}+3 \hat{k}$ and coplanar with the vectors $\hat{i}+\hat{j}+\hat{k}$ and $2 \hat{i}-\hat{j}-\hat{k}$ is

If $A, B, C, D$ are any four points in space, then $|\overrightarrow{AB} \times \overrightarrow{CD} + \overrightarrow{BC} \times \overrightarrow{AD} + \overrightarrow{CA} \times \overrightarrow{BD}|$ is equal to (where $\Delta$ denotes the area of $\Delta ABC$) (in $\Delta$)

If the position vectors of the vertices of a $\triangle ABC$ are $\vec{OA} = 3\hat{i} + \hat{j} + 2\hat{k}$,$\vec{OB} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{OC} = 2\hat{i} + 3\hat{j} + \hat{k}$,then the length of the altitude of $\triangle ABC$ drawn from $A$ is

Let $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\bar{b}=\hat{i}+\hat{j}$ and $\bar{c}$ be a vector such that $|\bar{c}-\bar{a}|=4$,$|(\bar{a} \times \bar{b}) \times \bar{c}|=3$ and the angle between $\bar{c}$ and $\bar{a} \times \bar{b}$ is $\frac{\pi}{6}$,then $\bar{a} \cdot \bar{c}$ is equal to