Let \(\vec{a},\text{ }\vec{b}\) be two vectors, and let \(P, Q\) and \(R\) be the points with position vectors \(\vec{a}…

Let \(\vec a,\vec b\) be the position vectors of \(P,Q\), and \(\vec a+\vec b\) be the position vector of \(R\).

Given \(|\vec a+\vec b|=\sqrt{21}\), so \(|\vec a+\vec b|^2=21\) ...(1)

\(|\vec a-\vec b|=3\), so \(|\vec a-\vec b|^2=9\) ...(2)

Also \(\vec a\) and \(\vec a-\vec b\) are perpendicular.

So \(\vec a\cdot(\vec a-\vec b)=0\)

\(\Rightarrow |\vec a|^2-\vec a\cdot\vec b=0\)

\(\Rightarrow \vec a\cdot\vec b=|\vec a|^2\) ...(3)

from (1), (2) and (3):

\(\vec a\cdot\vec b=|\vec a|^2=3\)

and \(|\vec b|^2=12\)

Area of triangle \(OPR\) is

\(\dfrac12|\vec a\times(\vec a+\vec b)|\)

\(=\dfrac12|\vec a\times\vec b|\)

Now \(|\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2\)

\(=3\cdot12-3^2=36-9=27\)

\(|\vec a\times\vec b|=3\sqrt3\)

Area \(=\dfrac12\cdot3\sqrt3=\dfrac{3\sqrt3}{2}\)