Let $a, b, c$ be unit vectors such that $a$ is perpendicular to the plane of $b$ and $c$. If the angle between $b$ and $c$ is $\frac{\pi}{3}$, then $|a + b + c| =$

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $|\overrightarrow{a}|=3, |\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=\sqrt{37}$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is:

Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c}+\vec{d}|=\sqrt{29}$ and $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$. If $\lambda_1, \lambda_2$ $(\lambda_1 > \lambda_2)$ are the possible values of $(\vec{c}+\vec{d}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k})$, then the equation $K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+(3K+\frac{\lambda_{2}}{2})y^{2}-8x+12y+\lambda_{2}=0$ represents a circle, for $K$ equal to:

Let $a, b, c$ be three vectors such that $a \neq 0$,$a \times b = 2a \times c$,$|a| = |c| = 1$,$|b| = 4$,and $|b \times c| = \sqrt{15}$. If $b - 2c = \lambda a$,then $\lambda$ equals:

$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = \dots$

Let $\overline{A}, \overline{B}, \overline{C}$ be vectors of lengths $3$ units,$4$ units,and $5$ units respectively. If $\overline{A}$ is perpendicular to $\overline{B}+\overline{C}$,$\overline{B}$ is perpendicular to $\overline{C}+\overline{A}$,and $\overline{C}$ is perpendicular to $\overline{A}+\overline{B}$,then the length of vector $\overline{A}+\overline{B}+\overline{C}$ is