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

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

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 $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = \cos \theta$. Then the maximum value of $\theta$ is,where $\theta \in [0, \pi]$.

The number of distinct real values of $\lambda$,for which the vectors $-\lambda^2 \hat{i}+\hat{j}+\hat{k}$,$\hat{i}-\lambda^2 \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}-\lambda^2 \hat{k}$ are coplanar,is

$A$ unit vector $\vec{e} = a \hat{i} + b \hat{j} + c \hat{k}$ is coplanar with the vectors $\hat{i} - 3 \hat{j} + 5 \hat{k}$ and $3 \hat{i} + \hat{j} - 5 \hat{k}$. If $\vec{e}$ is perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$,then $2 a^2 + 3 b^2 + 4 c^2 =$