The value of $\lambda$ for which points $A(2, 2, 1)$,$B(1, 1, 1)$,$C(-\lambda, 2, 1)$,and $D(3, 0, -1)$ are coplanar is $\lambda = $ ............

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:

The position vectors of the points $A, B, C$ and $D$ are $3 \hat{i}-2 \hat{j}-\hat{k}, 2 \hat{i}-3 \hat{j}+2 \hat{k}, \hat{i}-\hat{j}+2 \hat{k}$ and $4 \hat{i}-\hat{j}-\lambda \hat{k}$ respectively. If the points $A, B, C$ and $D$ lie on a plane,the value of $\lambda$ is

If $a = i + j + k$,$b = 4i + 3j + 4k$ and $c = i + \alpha j + \beta k$ are linearly dependent vectors and $|c| = \sqrt{3}$,then

If $\overline{a}, \overline{b}$ and $\overline{c}$ are unit coplanar vectors,then the scalar triple product $[2 \overline{a}-\overline{b}, 2 \overline{b}-\overline{c}, 2 \overline{c}-\overline{a}]$ has the value

Three vectors $\vec{a} = \hat{i} + \hat{j}$,$\vec{b} = \hat{j} + \hat{k}$,and $\vec{c} = \hat{k} + \hat{i}$ are given. If three unit vectors are drawn perpendicular to the three planes formed by these vectors,what is the volume of the parallelepiped formed by these unit vectors?