If $\overline{a}, \overline{b}$ and $\overline{c}$ are unit coplanar vectors,then the scalar triple product $[2 \overline{a}-\overline{b}, 2 \overline{b}-\overline{c}, 2 \overline{c}-\overline{a}]$ has the value

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:

If the four points $A(6,2,4)$,$B(1,3,5)$,$C(1,-2,3)$,and $D(6, k, 2)$ are coplanar,then $k=$

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:

If $a, b, c, d$ are $4$ vectors such that $a \cdot b = 0$, $|a \times c| = |a||c|$, and $|a \times d| = |a||d|$, then $[b c d] = $

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