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 = $ ............

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

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 $\vec{a} = \hat{i} + 2\hat{j} + 4\hat{k}$,$\vec{b} = \hat{i} + \lambda\hat{j} + 4\hat{k}$,and $\vec{c} = 2\hat{i} + 4\hat{j} + (\lambda^2 - 1)\hat{k}$ be coplanar vectors. Then the non-zero vector $\vec{a} \times \vec{c}$ is:

Let a vector $\vec{a}$ be coplanar with vectors $\vec{b}=2 \hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\vec{d}=3 \hat{i}+2 \hat{j}+6 \hat{k}$,and $|\vec{a}|=\sqrt{10}$. Then a possible value of $[\vec{a} \vec{b} \vec{c}]+[\vec{a} \vec{b} \vec{d}]+[\vec{a} \vec{c} \vec{d}]$ is equal to:

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $(\vec{a} - \lambda \vec{b}) \cdot (\vec{b} - 2\vec{c}) \times (\vec{c} + 2\vec{a}) = 0$,then $\lambda$ is equal to