The volume of the tetrahedron formed by the vectors $\vec{a}, \vec{b}, \vec{c}$ is $3$. Then the volume of the parallelepiped formed by the coterminous edges $\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}$ is:

If $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular vectors such that $|\bar{a}| = a, |\bar{b}| = b, |\bar{c}| = c$,then $[\bar{a} \bar{b} \bar{c}] = ......$

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector,assuming the question implies the vectors are linearly dependent or part of a coplanar set,find $\lambda$. Given the standard form of such problems,if we consider the vectors $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ to be coplanar with a reference vector,let us assume the third vector is $\hat{k}$. For these to be coplanar,the scalar triple product must be zero: $\left|\begin{array}{ccc} 1 & -3 & 1 \\ \lambda & 3 & 0 \\ 0 & 0 & 1 \end{array}\right| = 0$. Solving this,$\lambda$ is equal to:

If $3 \hat{i}+3 \hat{j}+\sqrt{3} \hat{k}$,$\hat{i}+\hat{k}$,and $\sqrt{3} \hat{i}+\sqrt{3} \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda$ is equal to

If the scalar triple product of the vectors $-3 \hat{i}+7 \hat{j}-3 \hat{k}$,$3 \hat{i}-7 \hat{j}+\lambda \hat{k}$ and $7 \hat{i}-5 \hat{j}-3 \hat{k}$ is $272$,then $\lambda = \ldots$

Find the volume of a tetrahedron whose vertices are given by the vectors $-i + j + k$,$i - j + k$,and $i + j - k$,with the fourth vertex being the origin.