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

If the area of the triangle with vertices $\hat{i}+y \hat{j}$,$\hat{i}+2 \hat{k}$,and $3 \hat{j}+\hat{k}$ is $\sqrt{6}$ sq. units,then the values of $y$ are

Let $\vec{a} = \vec{j} - \vec{k}$ and $\vec{c} = \vec{i} - \vec{j} - \vec{k}$. Find the vector $\vec{b}$ satisfying $\vec{a} \times \vec{b} + \vec{c} = 0$ and $\vec{a} \cdot \vec{b} = 3$.

If $\vec{u} = \vec{a} - \vec{b}$ and $\vec{v} = \vec{a} + \vec{b}$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\vec{u} \times \vec{v}| = ......$

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

If the vectors $\vec{a} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j}$ represent the diagonals of a parallelogram,then its area will be: