If $\vec{a} = 4\hat{i} - 2\hat{j} + \hat{k}$,$\vec{b} = 3\hat{i} + 2\hat{j} - \hat{k}$,and $\vec{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ represent the three coterminous edges of a parallelepiped,find its volume.

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and the points $P_1 = \lambda \vec{a}+3 \vec{b}-\vec{c}$,$P_2 = \vec{a}-\lambda \vec{b}+3 \vec{c}$,$P_3 = 3 \vec{a}+4 \vec{b}-\lambda \vec{c}$,and $P_4 = \vec{a}-6 \vec{b}+6 \vec{c}$ are coplanar,then one of the values of $\lambda$ 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 $\vec{c} = (0, 1, 0)$:

If the vectors $2\hat{i} - \hat{j} + \hat{k}$,$\hat{i} + 2\hat{j} - 3\hat{k}$,and $3\hat{i} + a\hat{j} + 5\hat{k}$ are coplanar,then the value of $a$ is:

If $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar vectors and $p, q$ are real numbers,then the equality $[3\vec{u}, p\vec{v}, p\vec{w}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w}, q\vec{v}, q\vec{u}] = 0$ holds for which of the following?

If the volume of a tetrahedron,whose vertices are $A(1, 2, 3)$,$B(-3, -1, 1)$,$C(2, 1, 3)$,and $D(-1, 2, x)$ is $\frac{11}{6}$ cubic units,then the value of $x$ is: