$A, B, C, D$ are any $4$ points and $|\overline{AB} \times \overline{CD} + \overline{BC} \times \overline{AD} + \overline{CA} \times \overline{BD}| = \lambda$ (Area of $\triangle ABC$). Then $\lambda = $

$A$ unit vector perpendicular to the vectors $4i - j + 3k$ and $-2i + j - 2k$ is

Let $a=2 \hat{i}-2 \hat{j}+\hat{k}$ and $b=-\hat{j}+\hat{k}$. If $c$ is a vector such that $a \cdot c=|c|$, $|c-a|=2 \sqrt{2}$, and the angle between $a \times b$ and $c$ is $\frac{\pi}{3}$, then $|(a \times b) \times c|=$

Find a unit vector coplanar with $i + 2j + k$ and $i + j + 2k$ and perpendicular to $i + j + k$.

Let $\vec{a}=6 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}$. If $\vec{c}$ is a vector such that $|\vec{c}| \geq 6$,$\vec{a} \cdot \vec{c}=6|\vec{c}|$,$|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^{\circ}$,then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to:

Find all vectors of magnitude $10 \sqrt{3}$ that are perpendicular to the plane of $\hat{i}+2 \hat{j}+\hat{k}$ and $-\hat{i}+3 \hat{j}+4 \hat{k}$.