If the position vector of A, B are $2\overset{\to }{\mathop{a}}\,-3\overset{\to }{\mathop{b}}\,$ and $3\overset{\to }{\mathop{a}}\,+2\overset{\to }{\mathop{b}}\,$, then the p.v. of C in \[\overset{\to }{\mathop{AB}}\,\]produced such that $\overset{\to }{\mathop{AC}}\,=2\overset{\to }{\mathop{AB}}\,$ is
a) $3\overset{\to }{\mathop{a}}\,+2\overset{\to }{\mathop{b}}\,$
b) $3\overset{\to }{\mathop{b}}\,-2\overset{\to }{\mathop{a}}\,$
c) \[4\overset{\to }{\mathop{a}}\,+7\overset{\to }{\mathop{b}}\,\]
d) \[5\overset{\to }{\mathop{b}}\,-2\overset{\to }{\mathop{a}}\,\]

Hint: We have position vectors of A and B as: $2\overset{\to }{\mathop{a}}\,-3\overset{\to }{\mathop{b}}\,$ and $3\overset{\to }{\mathop{a}}\,+2\overset{\to }{\mathop{b}}\,$. So, we can write vector $\overset{\to }{\mathop{AB}}\,$as: $\overset{\to }{\mathop{AB}}\,=\left( \overset{\to }{\mathop{B}}\,-\overset{\to }{\mathop{A}}\, \right)$. Similarly, we can write: $\overset{\to }{\mathop{AC}}\,=\left( \overset{\to }{\mathop{C}}\,-\overset{\to }{\mathop{A}}\, \right)$. As we know that: $\overset{\to }{\mathop{AC}}\,=2\overset{\to }{\mathop{AB}}\,$, so using this relation, we can find the position vectors of C.

Complete step-by-step solution:
We are given the position vector of A as $2\overset{\to }{\mathop{a}}\,-3\overset{\to }{\mathop{b}}\,$ and position vector of B as $3\overset{\to }{\mathop{a}}\,+2\overset{\to }{\mathop{b}}\,$ according to the question
Now we know that the vector from A to B that is $\overrightarrow{AB}$ , can be written as:
$\overset{\to }{\mathop{AB}}\,=\left( \overset{\to }{\mathop{B}}\,-\overset{\to }{\mathop{A}}\, \right)......(1)$
Now we will put the values of position vectors of A and B in equation (1), therefore we will get:
$\begin{align}
  & \overrightarrow{AB}=\left( \overrightarrow{B}-\overrightarrow{A} \right)=\left( 3\overrightarrow{a}+2\overrightarrow{b} \right)-\left( 2\overrightarrow{a}-3\overrightarrow{b} \right)=\overrightarrow{a}+5\overrightarrow{b}\text{ } \\
 & \overrightarrow{AB}=\overrightarrow{a}+5\overrightarrow{b}\text{ }.......\left( 2 \right) \\
\end{align}$

Now we can also write:
$\overset{\to }{\mathop{AC}}\,=\left( \overset{\to }{\mathop{C}}\,-\overset{\to }{\mathop{A}}\, \right)$
Now, we will put the position vector value of $\overrightarrow{A}$ , therefore we will get:
$\overset{\to }{\mathop{AC}}\,=\overset{\to }{\mathop{C}}\,-\left( 2\overset{\to }{\mathop{a}}\,-3\overset{\to }{\mathop{b}}\, \right)......(3)$
Now it is given in the question that: $\overset{\to }{\mathop{AC}}\,=2\overset{\to }{\mathop{AB}}\,$,
So, we will put the values from equation (2) and (3), we can write:
$\begin{align}
  & \overset{\to }{\mathop{C}}\,-\left( 2\overset{\to }{\mathop{a}}\,-3\overset{\to }{\mathop{b}}\, \right)=2\left( \overset{\to }{\mathop{a}}\,+5\overset{\to }{\mathop{b}}\, \right) \\
 & \overset{\to }{\mathop{C}}\,=\left( 2\overset{\to }{\mathop{a}}\,+10\overset{\to }{\mathop{b}}\, \right)+\left( 2\overset{\to }{\mathop{a}}\,-3\overset{\to }{\mathop{b}}\, \right) \\
 & \overset{\to }{\mathop{C}}\,=\left( 4\overset{\to }{\mathop{a}}\,+7\overset{\to }{\mathop{b}}\, \right) \\
\end{align}$
Hence the position vector of C is: $\overset{\to }{\mathop{C}}\,=\left( 4\overset{\to }{\mathop{a}}\,+7\overset{\to }{\mathop{b}}\, \right)$
Therefore, option (c) is the correct answer.

Note: In a 2D cartesian coordinate system, the position of a point is given in terms of x and y. The position of a point is given by (x, y).
A position vector relates that point to its origin. For a single position vector, the origin is where the x-axis and y-axis intersect; where x=0 and y=0 or (0, 0).
So, if the position of point A is (x, y), then the position vector of A is given by
$\overset{\to }{\mathop{OA}}\,=x\hat{i}+y\hat{j}$, where O is the origin and $\hat{i}$ and $\hat{j}$are unit vectors.
So, in short, it is a vector that starts at the origin, (0,0), and ends at position (x, y). It provides, both, the direction and magnitude. In this case, the magnitude is the displacement from (0,0) to (x, y).