If $c = 2 \lambda (a \times b) + 3 \mu (b \times a)$ where $a \times b \neq 0$ and $c \cdot (a \times b) = 0$,then:

If $a = 2i + j + 2k$ and $b = 5i - 3j + k$,then the projection of $b$ on $a$ is

If $\vec{a} = \hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{b} = 4\hat{i} - 2\hat{j} + 4\hat{k}$,then $(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b}) = \dots$

Let $\overrightarrow{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\overrightarrow{b} = 7\hat{i} + \hat{j} - 6\hat{k}$. If $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{r} \times \overrightarrow{b}$ and $\overrightarrow{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$,then $\overrightarrow{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k})$ is equal to:

If the vectors $\vec{BC} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{CD} = \hat{i} + 2\hat{j} - 2\hat{k}$ represent two adjacent sides of a parallelogram $ABCD$ and $\theta$ is the angle between its diagonals $\vec{AC}$ and $\vec{BD}$,then $\tan \theta =$

$A, B, C, D$ are any four points,then $\overrightarrow{AB} \cdot \overrightarrow{CD} + \overrightarrow{BC} \cdot \overrightarrow{AD} + \overrightarrow{CA} \cdot \overrightarrow{BD} = $