Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\vec{c} = x\hat{i} + (x-2)\hat{j} - \hat{k}$. If the vector $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$,then $x$ equals:

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 $a=2 \hat{i}-3 \hat{j}+5 \hat{k}$, $b=3 \hat{i}-4 \hat{j}+5 \hat{k}$ and $c=5 \hat{i}-3 \hat{j}-2 \hat{k}$, then the volume of the parallelepiped with coterminous edges $a+b$, $b+c$, $c+a$ is

Statement $1$: The vectors $\vec{a}, \vec{b}$ and $\vec{c}$ lie in the same plane if and only if $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.
Statement $2$: The vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if $\vec{u} \cdot \vec{v} = 0$,where $\vec{u} \times \vec{v}$ is a vector perpendicular to the plane of $\vec{u}$ and $\vec{v}$.

Let $x_0$ be the point of local maxima of $f(x) = \vec{a} \cdot (\vec{b} \times \vec{c})$, where $\vec{a} = x\hat{i} - 2\hat{j} + 3\hat{k}$, $\vec{b} = -2\hat{i} + x\hat{j} - \hat{k}$, and $\vec{c} = 7\hat{i} - 2\hat{j} + x\hat{k}$. Then the value of $\vec{a} \cdot \vec{c}$ at $x = x_0$ is:

Let $\overrightarrow{a}=\hat{i}-2 \hat{j}$,$\overrightarrow{b}=2 \hat{j}+3 \hat{k}$,$\overrightarrow{c}=p\hat{i}+q \hat{j}$ and $\overrightarrow{d}=p \hat{j}-q \hat{k}$ be four vectors. If $(\vec{a} \times \vec{b}) \cdot \vec{c}=3=(\vec{a} \times \vec{b}) \cdot \vec{d}$,then $3 p+q=$