The vectors $\overline{p}=\hat{i}+a \hat{j}+a^2 \hat{k}$,$\overline{q}=\hat{i}+b \hat{j}+b^2 \hat{k}$ and $\overline{r}=\hat{i}+c \hat{j}+c^2 \hat{k}$ are non-coplanar 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:

Let $\vec{p} = 2\hat{i} + 3\hat{j} + a\hat{k}$,$\vec{q} = b\hat{i} + 5\hat{j} - \hat{k}$,and $\vec{r} = \hat{i} + \hat{j} + 3\hat{k}$. If $\vec{p}, \vec{q}, \vec{r}$ are coplanar and $\vec{p} \cdot \vec{q} = 20$,then the ordered pair $(a, b)$ is:

If $[\bar{a} \bar{b} \bar{c}] \neq 0$,then $\frac{[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]}{[\bar{b} \bar{c} \bar{a}]}=$

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.

If $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$,$\vec{b}=\hat{i}+3 \hat{j}-\hat{k}$ and $\vec{c}=3 \hat{i}-\hat{j}-2 \hat{k}$,then the value of $\left|\begin{array}{lll}\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c}\end{array}\right|$ is:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+2\hat{k}$,$\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ and $\vec{c}$ is a linear combination of $\vec{a}$ and $\vec{b}$,then the value of $x$ is: