If $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{b} \times \vec{c} = \vec{a}$,and $a, b, c$ are the moduli of the vectors $\vec{a}, \vec{b}, \vec{c}$ respectively,then:

If $\vec{x}$ is a unit vector such that $\vec{x} \times (\hat{i} - 2\hat{j} + \hat{k}) = -\hat{i} + \hat{k}$,then $\vec{x}$ is:

Let $ABC$ be a triangle of area $15 \sqrt{2}$ and the vectors $\overrightarrow{AB}=\hat{i}+2 \hat{j}-7 \hat{k}$,$\overrightarrow{BC}=a \hat{i}+b \hat{j}+c \hat{k}$ and $\overrightarrow{AC}=6 \hat{i}+d \hat{j}-2 \hat{k}$,where $d>0$. Then the square of the length of the largest side of the triangle $ABC$ is:

Find the unit vector perpendicular to both vectors $\vec{a}$ and $\vec{b}$.

If $a = i - j$,$b = i + j$,$c = i + 3j + 5k$ and $n$ is a unit vector such that $b \cdot n = 0$ and $a \cdot n = 0$,then the value of $|c \cdot n|$ is equal to

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. If $\overline{c}$ is a vector such that $\overline{a} \cdot \overline{c}=|\overline{c}|$,$|\overline{c}-\overline{a}|=2 \sqrt{2}$ and the angle between $(\overline{a} \times \overline{b})$ and $\overline{c}$ is $\frac{\pi}{6}$,then $|(\overline{a} \times \overline{b}) \times \overline{c}|$ is