For how many distinct real values of $\lambda$ are the vectors $-\lambda^2 \hat{i} + \hat{j} + \hat{k}$,$\hat{i} - \lambda^2 \hat{j} + \hat{k}$,and $\hat{i} + \hat{j} - \lambda^2 \hat{k}$ coplanar?

If $a, b, c$ are non-negative distinct numbers and $a \hat{\imath}+a \hat{\jmath}+c \hat{k}$,$\hat{\imath}+\hat{k}$ and $c \hat{\imath}+c \hat{\jmath}+b \hat{k}$ are coplanar vectors,then

Let $a=\hat{i}+\hat{j}$, $b=\hat{j}+\hat{k}$ and $c=\hat{i}+\hat{k}$. If $d$ is a unit vector such that $a \cdot d=0$ and $b \cdot(c \times d)=0$, then $d=$

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:

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 $\overrightarrow{p}=(a+1) \hat{i}+a \hat{j}+a \hat{k}$,$\overrightarrow{q}=a \hat{i}+(a+1) \hat{j}+a \hat{k}$,and $\overrightarrow{r}=a \hat{i}+a \hat{j}+(a+1) \hat{k}$ $(a \in R)$ are coplanar and $3(\overrightarrow{p} \cdot \overrightarrow{q})^{2}-\lambda|\overrightarrow{r} \times \overrightarrow{q}|^{2}=0$,then the value of $\lambda$ is: