If the points $A(2,1,-1), B(0,-1,0), C(4,0,4)$ and $D(2,0,x)$ are coplanar,then $x=$

The position vectors of the points $A, B, C$ and $D$ are $3 \hat{i}-2 \hat{j}-\hat{k}, 2 \hat{i}-3 \hat{j}+2 \hat{k}, \hat{i}-\hat{j}+2 \hat{k}$ and $4 \hat{i}-\hat{j}-\lambda \hat{k}$ respectively. If the points $A, B, C$ and $D$ lie on a plane, the value of $\lambda$ is

If four distinct points with position vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are coplanar,then $[\vec{a} \vec{b} \vec{c}]$ is equal to

Let $a, b, c$ be three distinct real numbers,none equal to $1$. If the vectors $a \hat{i}+\hat{j}+\hat{k}$,$\hat{i}+b \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+ c \hat{k}$ are coplanar,then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is equal to

If $\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}+\hat{k}$,and $\vec{c}=3 \hat{i}-\hat{j}+2 \hat{k}$,then $\vec{a} \cdot(\vec{b} \times \vec{c})=$ . . . . . . .

If the vectors $\overrightarrow{a}=\hat{i}+a \hat{j}+a^{2} \hat{k}$,$\overrightarrow{b}=\hat{i}+b \hat{j}+b^{2} \hat{k}$ and $\overrightarrow{c}=\hat{i}+c \hat{j}+c^{2} \hat{k}$ are three non-coplanar vectors and $\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0$,then the value of $abc$ is