$A$ vector with magnitude of $3$ units,which is perpendicular to each of the vectors $\vec{a}=3 \hat{i}+\hat{j}-4 \hat{k}$ and $\vec{b}=6 \hat{i}+5 \hat{j}-2 \hat{k}$,is given by

Let $\overrightarrow{a}=3\hat{i}-\hat{j}+2\hat{k}$,$\overrightarrow{b}=\overrightarrow{a}\times(\hat{i}-2\hat{k})$ and $\overrightarrow{c}=\overrightarrow{b}\times\hat{k}$. Then the projection of $\overrightarrow{c}-2\hat{j}$ on $\overrightarrow{a}$ is:

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

The number of vectors of unit length perpendicular to vectors $a = (1, 1, 0)$ and $b = (0, 1, 1)$ is

Let the vectors $\overline{a}, \overline{b}, \overline{c}$ and $\overline{d}$ be such that $(\overline{a} \times \overline{b}) \times(\overline{c} \times \overline{d})=\overline{0}$. Let $P_1$ and $P_2$ be the planes determined by the pair of vectors $\overline{a}, \overline{b}$ and $\overline{c}, \overline{d}$ respectively,then the angle between $P_1$ and $P_2$ is

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{j}$, $\overrightarrow{c}=\hat{i}$ and $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$, then $\lambda+\mu$ is equal to: