Let $\overrightarrow{a},\overrightarrow{b}$ be two vectors, and let $P,Q$ and $R$ be the points with position vectors $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{a}+\overrightarrow{b}$, respectively, with respect to the origin $O$. If $|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{21}$, $|\overrightarrow{a}-\overrightarrow{b}|=3$, and $\overrightarrow{a}$ and $(\overrightarrow{a}-\overrightarrow{b})$ are perpendicular to each other, then the area of the triangle $OPR$ is

The correct option is $\frac{3\sqrt{3}}{2}$.

Let us analyze the given information step-by-step:

1. The position vectors of points P, Q, and R relative to the origin O are given as:

$\overrightarrow{OP}=\overrightarrow{a}$

$\overrightarrow{OQ}=\overrightarrow{b}$

$\overrightarrow{OR}=\overrightarrow{a}+\overrightarrow{b}$

2. We are given the magnitudes:

$|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{21}$

$|\overrightarrow{a}-\overrightarrow{b}|=3$

3. We are also told that $\overrightarrow{a}$ and $(\overrightarrow{a}-\overrightarrow{b})$ are perpendicular to each other. Therefore, their dot product is zero:

$\overrightarrow{a}·(\overrightarrow{a}-\overrightarrow{b})=0$

$|\overrightarrow{a}{|}^2-\overrightarrow{a}·\overrightarrow{b}=0\Rightarrow \overrightarrow{a}·\overrightarrow{b}=|\overrightarrow{a}{|}^2$

4. Now, let us square both given magnitude equations:

$|\overrightarrow{a}+\overrightarrow{b}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+2(\overrightarrow{a}·\overrightarrow{b})=21$

$|\overrightarrow{a}-\overrightarrow{b}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2-2(\overrightarrow{a}·\overrightarrow{b})=9$

5. Subtracting the second equation from the first equation gives:

$4(\overrightarrow{a}·\overrightarrow{b})=21-9=12$

$\stackrel{·}{\overrightarrow{a}}\overrightarrow{b}=3$

Since $\overrightarrow{a}·\overrightarrow{b}=|\overrightarrow{a}{|}^2$, we have:

$|\overrightarrow{a}{|}^2=3\Rightarrow |\overrightarrow{a}|=\sqrt{3}$

6. Adding the two squared equations gives:

$2(|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2)=21+9=30$

$|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2=15$

Substituting $|\overrightarrow{a}{|}^2=3$:

$3+|\overrightarrow{b}{|}^2=15\Rightarrow |\overrightarrow{b}{|}^2=12\Rightarrow |\overrightarrow{b}|=2\sqrt{3}$

7. The area of triangle OPR formed by vectors $\overrightarrow{OP}=\overrightarrow{a}$ and $\overrightarrow{OR}=\overrightarrow{a}+\overrightarrow{b}$ is given by:

$\text{Area}=\frac{1}{2}|\overrightarrow{OP}\times \overrightarrow{OR}|$

$\text{Area}=\frac{1}{2}|\overrightarrow{a}\times (\overrightarrow{a}+\overrightarrow{b})|$

Since $\overrightarrow{a}\times \overrightarrow{a}=\overrightarrow{0}$, this simplifies to:

$\text{Area}=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|$

8. Using Lagrange's identity, we can find $|\overrightarrow{a}\times \overrightarrow{b}|$:

$|\overrightarrow{a}\times \overrightarrow{b}{|}^2=|\overrightarrow{a}{|}^2|\overrightarrow{b}{|}^2-(\overrightarrow{a}·\overrightarrow{b}{)}^2$

$|\overrightarrow{a}\times \overrightarrow{b}{|}^2=(3\left)\right(12)-(3{)}^2=36-9=27$

$|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{27}=3\sqrt{3}$

9. Substituting this into the area formula gives:

$\text{Area of}\mathrm{\Delta }OPR=\frac{1}{2}\left(3\sqrt{3}\right)=\frac{3\sqrt{3}}{2}$