If $\overline{e_1}, \overline{e_2}$ are two non-collinear unit vectors such that $|\overline{e_1}+\overline{e_2}|=\sqrt{3}$,then $(2 \overline{e_1}-5 \overline{e_2}) \cdot (3 \overline{e_1}+\overline{e_2}) = $

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 $e$ is a unit vector perpendicular to the plane determined by the points $2 \hat{i}+\hat{j}+\hat{k}$,$\hat{i}-\hat{j}+\hat{k}$ and $-\hat{i}+\hat{j}-\hat{k}$. If $a=2 \hat{i}-3 \hat{j}+6 \hat{k}$,then the projection vector of $a$ on $e$ is

If $a, b, c$ are unit vectors such that $a + b + c = 0,$ then $a \cdot b + b \cdot c + c \cdot a = $

If $a \neq 0, b \neq 0$ and $|a + b| = |a - b|,$ then the vectors $a$ and $b$ are

Let $\lambda \in \mathbb{Z}$,$\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$. Let $\vec{c}$ be a vector such that $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = \vec{0}$,$\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$. Then $|\vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$ is equal to