If $a, b, c$ are unit vectors satisfying the relation $a+b+\sqrt{3} c=0$, then the angle between $a$ and $b$ is

Let $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}$,$\vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$,and $\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}$ and $\vec{r} \cdot (\vec{b}-\vec{c})=0$,then $\frac{|593 \vec{r}+67 \vec{a}|^2}{(593)^2}$ is equal to:

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

If $a, b, c, d$ are the position vectors of the points $A, B, C$ and $D$ respectively,referred to the same origin $O$,such that no three of these points are collinear and $a + c = b + d$,then the quadrilateral $ABCD$ is a

Let $\vec{a}, \vec{b}, \vec{c}$ be vectors of lengths $3, 4, 5$ respectively. Let $\vec{a}$ be perpendicular to $\vec{b}+\vec{c}$,$\vec{b}$ be perpendicular to $\vec{c}+\vec{a}$,and $\vec{c}$ be perpendicular to $\vec{a}+\vec{b}$. Then the length of vector $\vec{a}+\vec{b}+\vec{c}$ is:

If $\theta$ is the angle between the vectors $\bar{a}$ and $\bar{b}$ where $|\bar{a}|=4, |\bar{b}|=3$ and $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)$,then $|(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b})|^2+4(\bar{a} \cdot \bar{b})^2$ has the value