If $[\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}] = \lambda [\vec{a}, \vec{b}, \vec{c}]^2$,then $\lambda$ is equal to:

If the four points,whose position vectors are $3 \hat{i} - 4 \hat{j} + 2 \hat{k}$,$\hat{i} + 2 \hat{j} - \hat{k}$,$-2 \hat{i} - \hat{j} + 3 \hat{k}$,and $5 \hat{i} - 2 \alpha \hat{j} + 4 \hat{k}$ are coplanar,then $\alpha$ is equal to

The volume of a tetrahedron whose vertices are $4 \hat{i}+5 \hat{j}+\hat{k}$, $-\hat{j}+\hat{k}$, $3 \hat{i}+9 \hat{j}+4 \hat{k}$ and $-2 \hat{i}+4 \hat{j}+4 \hat{k}$ is (in cubic units)

If $\bar{a}=\hat{i}+\hat{j}+\hat{k}$,$\bar{b}=4\hat{i}+3\hat{j}+4\hat{k}$,and $\bar{c}=\hat{i}+\alpha\hat{j}+\beta\hat{k}$ are linearly dependent vectors and $|\bar{c}|=\sqrt{3}$,then the values of $\alpha$ and $\beta$ are respectively.

If the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ are coplanar, then $x=$

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i}$, and $\vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}$ with $c_1 = 1$ and $c_2 = 2$, then find the value of $c_3$ such that $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar.