Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$, and $\overline{\mathrm{c}}$ be three non-zero vectors such that no two of these are collinear. If the vector $\bar{a}+2 \bar{b}$ is collinear with $\bar{c}$ and $\bar{b}+3 \bar{c}$ is collinear with $\overline{\mathrm{a}}$, then $\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}+6 \overline{\mathrm{c}}$ equals

If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non-coplanar unit vectors such that $\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\frac{(\overline{\mathrm{b}}+\overline{\mathrm{c}})}{\sqrt{2}}$ then the angle between $\overline{\mathrm{a}}$ and $\bar{b}$ is

The number of unit vectors perpendicular to $\overline{\mathrm{a}}=(1,1,0)$ and $\overline{\mathrm{b}}=(0,1,1)$ is

If the vectors $\overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+\mu \hat{\mathrm{k}}$ are mutually orthogonal, then $(\lambda, \mu) \equiv$