$A$ unit vector coplanar with $i+j+3k$ and $i+3j+k$ and perpendicular to $i+j+k$ is

If $a, b, c$ are any three vectors and their reciprocal vectors are $a^{-1}, b^{-1}, c^{-1}$ such that $[a, b, c] \neq 0$,then $[a^{-1}, b^{-1}, c^{-1}]$ is equal to:

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 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:

If the four points $2\vec{a} + 3\vec{b} - \vec{c}$,$\vec{a} - 2\vec{b} + 3\vec{c}$,$3\vec{a} + 4\vec{b} - 2\vec{c}$,and $\vec{a} - \lambda\vec{b} - 6\vec{c}$ are coplanar,find the value of $\lambda$.

If $a, b,$ and $c$ are coplanar unit vectors,find the value of the scalar triple product $[2a - b, 2b - c, 2c - a]$.