Step 1: Understanding the Question:
The problem involves vector geometry where we need to find the area of a triangle formed by the origin and two specific position vectors. We are given magnitudes of the sum and difference of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, along with a perpendicularity condition.

Step 2: Key Formula or Approach:
• Use the parallelogram law for magnitudes: ${|\overrightarrow{u}+\overrightarrow{v}|}^2={|\overrightarrow{u}|}^2+{|\overrightarrow{v}|}^2+2\overrightarrow{u}·\overrightarrow{v}$.
• Area of triangle $OPR$ with vertices $O\left(\overrightarrow{0}\right),P\left(\overrightarrow{a}\right),R(\overrightarrow{a}+\overrightarrow{b})$ is given by $\frac{1}{2}|\overrightarrow{OP}\times \overrightarrow{OR}|$.
• Dot product of perpendicular vectors is zero.

Step 3: Detailed Explanation:
• Given $|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{21}{ \Rightarrow |\overrightarrow{a}|}^2+{|\overrightarrow{b}|}^2+2\stackrel{}{a}·\overrightarrow{b}=21$. (Equation 1)
• Given $|\overrightarrow{a}-\overrightarrow{b}|=3\Rightarrow {|\overrightarrow{a}|}^2+{|\overrightarrow{b}|}^2-2\overrightarrow{a}·\overrightarrow{b}=9$. (Equation 2)
• Subtracting Equation 2 from Equation 1: $4\overrightarrow{a}·\overrightarrow{b}=12\Rightarrow \overrightarrow{a}·\overrightarrow{b}=3$.
• Since $\overrightarrow{a}$ and $(\overrightarrow{a}-\overrightarrow{b})$ are perpendicular, $\overrightarrow{a}·(\overrightarrow{a}-\overrightarrow{b})=0$.
$\overrightarrow{a}·\overrightarrow{a}-\overrightarrow{a}·\overrightarrow{b}=0\Rightarrow {|\overrightarrow{a}|}^2=\overrightarrow{a}·\overrightarrow{b}=3$.
• From Equation 2: $3+{|\overrightarrow{b}|}^2-2\left(3\right)=9\Rightarrow {|\overrightarrow{b}|}^2-3=9\Rightarrow {|\overrightarrow{b}|}^2=12$.
• The area of triangle $OPR$ is:
Area = $\frac{1}{2}|\overrightarrow{a}\times (\overrightarrow{a}+\overrightarrow{b})|=\frac{1}{2}|(\overrightarrow{a}\times \overrightarrow{a})+(\overrightarrow{a}\times \overrightarrow{b})|=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|$.
• Using the identity ${|\overrightarrow{a}\times \overrightarrow{b}|}^2={|\overrightarrow{a}|}^2{|\overrightarrow{b}|}^2-{(\overrightarrow{a}·\overrightarrow{b})}^2$:
${|\overrightarrow{a}\times \overrightarrow{b}|}^2=\left(3\right)\left(12\right)-{\left(3\right)}^2=36-9=27\Rightarrow |\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{27}=3\sqrt{3}$.
• Therefore, Area = $\frac{1}{2}\left(3\sqrt{3}\right)=\frac{3\sqrt{3}}{2}$.

Step 4: Final Answer:
The area of triangle $OPR$ is $\frac{3\sqrt{3}}{2}$ square units.

Quick Tip: For any triangle with vertices A, B, C, the area can be calculated using the cross product of any two vectors formed by the vertices, such as $\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$. Remember that $\overrightarrow{a}\times \overrightarrow{a}=0$ simplifies such expressions significantly.