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:

$\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-coplanar vectors such that $\overrightarrow{P} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}$,$\overrightarrow{Q} = 4\overrightarrow{a} + 3\overrightarrow{b} + 4\overrightarrow{c}$,and $\overrightarrow{R} = \overrightarrow{a} + \alpha\overrightarrow{b} + \beta\overrightarrow{c}$ are linearly dependent vectors. Then,the number of possible values of $\alpha$ is:

The vector $c \cdot (b+c) \times (a+b+c)$ is equal to

Let $\vec{\lambda} = x\vec{a} + y\vec{b} + z\vec{c}$ and $\vec{\lambda} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = 2(x + y + z)$ (where $x + y + z \neq 0$),then the scalar triple product $[\vec{a} \, \vec{b} \, \vec{c}]$ is:

If the volume of the tetrahedron whose coterminous edges are given by the vectors $\vec{a} = -2\hat{i} + 3\hat{j} - 3\hat{k}$, $\vec{b} = 4\hat{i} + 5\hat{j} + (\lambda - 10)\hat{k}$, and $\vec{c} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ is $11$ cubic units, then the sum of the possible values of $\lambda$ is

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points $A, B, C$ and $D$ be $\vec{a}-\vec{b}+\vec{c}$,$\lambda \vec{a}-3 \vec{b}+4 \vec{c}$,$-\vec{a}+2 \vec{b}-3 \vec{c}$ and $2 \vec{a}-4 \vec{b}+6 \vec{c}$ respectively. If $\overrightarrow{AB}$,$\overrightarrow{AC}$ and $\overrightarrow{AD}$ are coplanar,then $\lambda$ is :