If $|a| = 3$ and $|b| = 4$,then a value of $\lambda$ for which $a + \lambda b$ is perpendicular to $a - \lambda b$ 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:

If $P=(0,1,2), Q=(4,-2,1)$ and $O=(0,0,0)$,then $\angle POQ=$

$7 \bar{i}-4 \bar{j}+7 \bar{k}, \bar{i}-6 \bar{j}+10 \bar{k}, -\bar{i}-3 \bar{j}+4 \bar{k}, 5 \bar{i}-\bar{j}+\bar{k}$ are the position vectors of the points $A, B, C, D$ respectively. If $p \bar{i}+q \bar{j}+r \bar{k}$ is the position vector of the point of intersection of the diagonals of the quadrilateral $ABCD$, then $p+q+r=$

If $a$ is a unit vector, then $|a \times \hat{i}|^2+|a \times \hat{j}|^2+|a \times \hat{k}|^2=$

The moment of the force $\overrightarrow{F}$ acting at a point $P$,about the point $C$ is: