If $35 \hat{i}+14 \hat{j}-77 \hat{k}$,$2 \hat{i}+7 \hat{j}+5 \hat{k}$ and $5 \hat{i}+2 \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda$ is equal to

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:

Let three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ be such that $\vec{a} \times \vec{b} = \vec{c}$,$\vec{b} \times \vec{c} = \vec{a}$ and $|\vec{a}| = 2$. Then which one of the following is not true?

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 $(1,5,35), (7,5,5), (1, \lambda, 7)$ and $(2 \lambda, 1, 2)$ are coplanar,then the sum of all possible values of $\lambda$ is

Volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is $12$ cubic units. The volume of a tetrahedron with coterminous edges $\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{a} + \vec{b} - \vec{c}$ will be ............. cubic units.