$a \cdot [(b + c) \times (a + b + c)]$ is equal to

The sum of all values of $\alpha$,for which the points whose position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$,$2 \hat{i}-3 \hat{j}+4 \hat{k}$,$(\alpha+1) \hat{i}+2 \hat{k}$ and $9 \hat{i}+(\alpha-8) \hat{j}+6 \hat{k}$ are coplanar,is equal to

If $x, y$ and $z$ are non-zero real numbers and $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ are such that $\vec{a} \times \vec{b}=z \hat{i}-3 \hat{j}+xy \hat{k}$ is not given,but $\vec{a} \times \vec{b}=6 \hat{i}-3 \hat{j}+\hat{k}$ is given as $z \hat{i}-3 \hat{j}+\hat{k}$,then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is equal to:

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 $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ at $x=x_{0}$ is:

If the vectors $2 \hat{i}-\hat{j}-\hat{k}$,$\hat{i}+2 \hat{j}-3 \hat{k}$,and $3 \hat{i}+\lambda \hat{j}+5 \hat{k}$ are coplanar,then the value of $\lambda$ is

Let $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{B} = \hat{i}$,and $\overrightarrow{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$. If $C_2 = -1$ and $C_3 = 1$,then to make the three vectors coplanar: