If the vectors $4i+11j+mk$,$7i+2j+6k$,and $i+5j+4k$ are coplanar,then $m$ is

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector,let us assume the vectors are $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$,$\vec{b} = \lambda \hat{i}+3 \hat{j}$,and we consider the standard basis vectors or a third vector to define coplanarity. However,if the question implies these two vectors are coplanar with the origin or a specific plane,we evaluate the scalar triple product. Given the standard interpretation of such problems,if $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ are coplanar with $\vec{c} = \hat{j}$,then the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 0$. Solving for $\lambda$ where $\vec{a} = (1, -3, 1)$,$\vec{b} = (\lambda, 3, 0)$,and $\vec{c} = (0, 1, 0)$:

$[(a \times b) \times (b \times c), (b \times c) \times (c \times a), (c \times a) \times (a \times b)] = \,$

Three concurrent edges $OA, OB, OC$ of a parallelepiped are represented by three vectors $2i + j - k$,$i + 2j + 3k$,and $-3i - j + k$. The volume of the solid so formed in cubic units is:

If four distinct points with position vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are coplanar,then $[\vec{a} \vec{b} \vec{c}]$ is equal to

If $x, y$ and $z$ are non-zero real numbers and $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ are such that $\vec{a} \times \vec{b}=z \hat{i}-3 \hat{j}+xy \hat{k}$ is not given,but $\vec{a} \times \vec{b}=6 \hat{i}-3 \hat{j}+\hat{k}$ is given as $z \hat{i}-3 \hat{j}+\hat{k}$,then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is equal to: