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

$\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$, $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}+\hat{j}-\hat{k}$ are three vectors. If $\vec{d}$ is a normal to the plane of $\vec{a}$ and $\vec{b}$ and $\vec{d} \cdot \vec{c}=2$, then $|\vec{d}|=$

If $a = 2 \hat{i} + 3 \hat{j} - \hat{k}$,$b = \hat{i} + 2 \hat{j} - 5 \hat{k}$,and $c = 3 \hat{i} + 5 \hat{j} - \hat{k}$,then a vector perpendicular to $a$ and in the plane containing $b$ and $c$ is:

Find a vector that is perpendicular to both vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + \hat{j}$.

If $\bar{a}=\hat{j}-\hat{k}$ and $\bar{c}=\hat{i}-\hat{j}-\hat{k}$,then the vector $\bar{b}$ satisfying $\bar{a} \times \bar{b}+\bar{c}=\vec{0}$ and $\bar{a} \cdot \bar{b}=3$ is

$A, B, C, D$ 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 = $