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:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=2\hat{i}+\lambda\hat{j}+\hat{k}$,$\vec{c}=\hat{i}-\hat{j}+4\hat{k}$ and $\vec{a} \cdot (\vec{b} \times \vec{c}) = 10$,then $\lambda$ is equal to

Let $a, b$ and $c$ be distinct non-negative numbers. If the vectors $a\hat{i} + a\hat{j} + c\hat{k}$,$\hat{i} + \hat{k}$ and $c\hat{i} + c\hat{j} + b\hat{k}$ are coplanar,then $c = \dots$

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-coplanar unit vectors such that the angle between every pair of them is $\frac{\pi}{3}$. If $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = p \vec{a} + q \vec{b} + r \vec{c}$,where $p, q$ and $r$ are scalars,then the value of $\frac{p^2 + 2q^2 + r^2}{q^2}$ is

Consider the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-6 \hat{k}$, $\vec{b}=6 \hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{c}=3 \hat{i}-6 \hat{j}-2 \hat{k}$.
Assertion $(A):$ The three vectors do not form a triangle.
Reason $(R):$ The three vectors are non-coplanar.
The correct option among the following is:

If $\vec{OA}=6 \hat{i}+3 \hat{j}-4 \hat{k}$, $\vec{OB}=2 \hat{j}+\hat{k}$, and $\vec{OC}=5 \hat{i}-\hat{j}+2 \hat{k}$ are the coterminous edges of a parallelepiped, then the height of the parallelepiped drawn from the vertex $A$ is