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

If the vectors $2 \bar{i} + 4 \bar{j} - 3 \bar{k}$,$-\bar{i} + 2 \bar{j} + 3 \bar{k}$,and $p \bar{i} - 2 \bar{j} + \bar{k}$ are coplanar,then the unit vector in the direction of the vector $9p \bar{i} - 4 \bar{j} + 4 \bar{k}$ is

Consider the set of eight vectors $V=\{a \hat{i}+b \hat{j}+c \hat{k}: a, b, c \in\{-1,1\}\}$. Three non-coplanar vectors can be chosen from $V$ in $2^p$ ways. Then $p$ is

Let the position vectors of the points $A, B, C$ and $D$ be $5\hat{i}+5\hat{j}+2\lambda\hat{k}$,$\hat{i}+2\hat{j}+3\hat{k}$,$-2\hat{i}+\lambda\hat{j}+4\hat{k}$ and $-\hat{i}+5\hat{j}+6\hat{k}$. Let the set $S = \{\lambda \in \mathbb{R} : \text{The points } A, B, C \text{ and } D \text{ are coplanar}\}$. Then $\sum_{\lambda \in S}(\lambda+2)^2$ is equal to

If the given vectors $(-bc, b^2 + bc, c^2 + bc)$,$(a^2 + ac, -ac, c^2 + ac)$ and $(a^2 + ab, b^2 + ab, -ab)$ are coplanar,where none of $a, b$ and $c$ is zero,then:

Let $a, b$ and $c$ be three vectors. Then the scalar triple product $[a, b, c]$ is equal to: