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

Let $\vec{a}=\hat{i}+2 \hat{j}-2 \hat{k}$ and $\vec{b}=2 \hat{i}-\hat{j}-2 \hat{k}$ be two vectors. If the orthogonal projection vector of $\vec{a}$ on $\vec{b}$ is $\vec{x}$ and the orthogonal projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{y}$,then find $|\vec{x}-\vec{y}|$.

If $a=2 \hat{i}+\hat{j}-3 \hat{k}$ and $b=3 \hat{i}-\hat{j}+2 \hat{k}$,then find the angle between the vectors $2 a+b$ and $a+2 b$.

If $a=\hat{i}+2 \hat{j}+3 \hat{k}$,$b=2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $c$ is a vector perpendicular to $b$,then $\left\{\frac{a \cdot(b \times c)}{|b \times c|^2}\right\}(b \times c)+\left\{\frac{a \cdot b}{|b|^2}\right\} b+\left\{\frac{a \cdot c}{|c|^2}\right\} c$ is equal to:

Suppose $\vec{a}+\vec{b}+\vec{c}=0$,$|\vec{a}|=3$,$|\vec{b}|=5$,$|\vec{c}|=7$. Then the angle between $\vec{a}$ and $\vec{b}$ 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: