If $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + \lambda \hat{j} + \mu \hat{k}) = \vec{0}$,then $\lambda$ and $\mu$ are respectively:

If $\overline{a}=\hat{i}+\hat{j}+\hat{k}$,$\overline{a} \cdot \overline{b}=1$ and $\overline{a} \times \overline{b}=\hat{j}-\hat{k}$,then $\overline{b}$ is

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

$\text{If } \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k} \text{ and } \vec{c} = \hat{i} - \hat{j} \text{ and if } 6\hat{i} + 2\hat{j} + 3\hat{k} = \lambda_1(\vec{a} \times \vec{b}) + \lambda_2(\vec{b} \times \vec{c}) + \lambda_3(\vec{c} \times \vec{a}), \text{ then } (\lambda_1, \lambda_2, \lambda_3) = $

For any two vectors $\vec{a}$ and $\vec{b}$,$|\vec{a} \times \vec{b}|^2$ is equal to

If $a$ and $b$ are two non-zero perpendicular vectors,then a vector $y$ satisfying the equations $a \cdot y = c$ (where $c$ is a scalar) and $a \times y = b$ is