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 . . . . . . .

Consider the following Assertion $(A)$ and Reason $(R)$:
Assertion $(A)$: The two lines $\bar{r}=\bar{a}+t(\bar{b})$ and $\bar{r}=\bar{b}+s(\bar{a})$ intersect each other.
Reason $(R)$: The shortest distance between the lines $\bar{r}=\bar{p}+t(\bar{q})$ and $\bar{r}=\bar{c}+s(\bar{d})$ is equal to the length of the projection of the vector $(\bar{p}-\bar{c})$ on $(\bar{q} \times \bar{d})$.
The correct answer is:

In a quadrilateral $ABCD$,if $P$ and $Q$ are the midpoints of $\overline{BC}$ and $\overline{AD}$ respectively,then $\vec{AB} + \vec{DC} = \dots$

Let $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ be vectors such that $\bar{a} \times \bar{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\bar{c} \times \bar{d} = 3\hat{i} + 2\hat{j} + \lambda\hat{k}$. If $\begin{vmatrix} \bar{a} \cdot \bar{c} & \bar{b} \cdot \bar{c} \\ \bar{a} \cdot \bar{d} & \bar{b} \cdot \bar{d} \end{vmatrix} = 0$,then find the value of $\lambda$.

The vector $\vec{a} = (\alpha, 2, \beta)$ lies in the plane of the vectors $\vec{b} = (1, 1, 0)$ and $\vec{c} = (0, 1, 1)$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives the possible values of $\alpha$ and $\beta$?

If $a=2 \hat{i}+\hat{k}$,$b=\hat{i}+\hat{j}+\hat{k}$,and $c=4 \hat{i}-3 \hat{j}+7 \hat{k}$,then the vector $r$ satisfying $r \times b=c \times b$ and $r \cdot a=0$ is