Let $a = \sin^2 x \hat{i} + \cos^2 x \hat{j} + \hat{k}$, where $x \in R$. If the pairs of vectors $(a, \hat{i})$, $(a, \hat{j})$, and $(a, \hat{k})$ are adjacent sides of $3$ distinct parallelograms and $A$ is the sum of the squares of the areas of these parallelograms, then $A$ lies in the interval

If $a, b, c$ are distinct real numbers and $P, Q, R$ are three points whose position vectors are respectively $a \hat{i}+b \hat{j}+c \hat{k}$, $b \hat{i}+c \hat{j}+a \hat{k}$ and $c \hat{i}+a \hat{j}+b \hat{k}$, then $\angle Q P R=$

If $a \cdot i = a \cdot (i + j) = a \cdot (i + j + k)$, then $a = $

If $4\hat{i}+\hat{j}-\hat{k}$ and $3\hat{i}+m\hat{j}+2\hat{k}$ are at a right angle,then $m = $

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} - \hat{k}$,$\overrightarrow{b} = \hat{i} - \hat{j}$ and $\overrightarrow{c} = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$ and $\overrightarrow{r} \cdot \overrightarrow{b} = 0$,then $\overrightarrow{r} \cdot \overrightarrow{a}$ is equal to ...........

Let $PQR$ be a triangle such that $\overrightarrow{PQ}=-2\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{PR}=a\hat{i}+b\hat{j}-4\hat{k}$, where $a, b \in \mathbb{Z}$. Let $S$ be the point on $QR$, which is equidistant from the lines $PQ$ and $PR$. If $|\overrightarrow{PR}|=9$ and $\overrightarrow{PS}=\hat{i}-7\hat{j}+2\hat{k}$, then the value of $3a-4b$ is . . . . . . .