Three vectors $\hat{i}-\hat{k}$,$\lambda \hat{i}+\hat{j}+(1-\lambda) \hat{k}$,and $\mu \hat{i}+\lambda \hat{j}+(1+\lambda-\mu) \hat{k}$ represent the coterminous edges of a parallelepiped. The volume of the parallelepiped depends on:

Consider the vectors $\vec{x}=\hat{i}+2\hat{j}+3\hat{k}$,$\vec{y}=2\hat{i}+3\hat{j}+\hat{k}$,and $\vec{z}=3\hat{i}+\hat{j}+2\hat{k}$. For two distinct positive real numbers $\alpha$ and $\beta$,define $\vec{X}=\alpha\vec{x}+\beta\vec{y}-\vec{z}$,$\vec{Y}=\alpha\vec{y}+\beta\vec{z}-\vec{x}$,and $\vec{Z}=\alpha\vec{z}+\beta\vec{x}-\vec{y}$. If the vectors $\vec{X}, \vec{Y}$,and $\vec{Z}$ lie in a plane,the value of $\alpha+\beta-3$ is $....$.

If $a, b, c$ are distinct positive numbers and vectors $a \hat{\imath} + a \hat{\jmath} + c \hat{k}$,$\hat{\imath} + \hat{k}$,and $c \hat{\imath} + c \hat{\jmath} + b \hat{k}$ lie in a plane,then

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and the points $P_1 = \lambda \vec{a}+3 \vec{b}-\vec{c}$, $P_2 = \vec{a}-\lambda \vec{b}+3 \vec{c}$, $P_3 = 3 \vec{a}+4 \vec{b}-\lambda \vec{c}$, and $P_4 = \vec{a}-6 \vec{b}+6 \vec{c}$ are coplanar, then one of the values of $\lambda$ is

Let $\vec{a}=\hat{i}-\alpha \hat{j}+\beta \hat{k}$,$\vec{b}=3 \hat{i}+\beta \hat{j}-\alpha \hat{k}$ and $\vec{c}=-\alpha \hat{i}-2 \hat{j}+\hat{k}$,where $\alpha$ and $\beta$ are integers. If $\vec{a} \cdot \vec{b}=-1$ and $\vec{b} \cdot \vec{c}=10$,then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to $.....$

If $(2,3,9), (5,2,1), (1, \lambda, 8)$ and $(\lambda, 2,3)$ are coplanar,then the product of all possible values of $\lambda$ is.