Let a , b be two vectors, and let P, Q and R be the points with...

Given $|\vec{a} + \vec{b}| = \sqrt{21}$ and $|\vec{a} - \vec{b}| = 3$.

Squaring both equations:

$|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 21$

$|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 9$

Subtracting the second equation from the first gives:

$4\vec{a} \cdot \vec{b} = 12 \Rightarrow \vec{a} \cdot \vec{b} = 3$

Adding the two equations gives:

$2(|\vec{a}|^2 + |\vec{b}|^2) = 30 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 = 15$

Since $\vec{a}$ and $(\vec{a} - \vec{b})$ are perpendicular, their dot product is zero:

$\vec{a} \cdot (\vec{a} - \vec{b}) = 0 \Rightarrow |\vec{a}|^2 - \vec{a} \cdot \vec{b} = 0 \Rightarrow |\vec{a}|^2 = \vec{a} \cdot \vec{b} = 3$

Substituting $|\vec{a}|^2 = 3$ into $|\vec{a}|^2 + |\vec{b}|^2 = 15$ gives:

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

Using Lagrange's identity to find $|\vec{a} \times \vec{b}|$:

$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$

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

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

The position vectors of $P$ and $R$ are $\vec{a}$ and $\vec{a} + \vec{b}$ respectively. The area of triangle $OPR$ is:

$\dfrac{1}{2} |\vec{OP} \times \vec{OR}| = \dfrac{1}{2} |\vec{a} \times (\vec{a} + \vec{b})| = \dfrac{1}{2} |\vec{a} \times \vec{a} + \vec{a} \times \vec{b}| = \dfrac{1}{2} |\vec{a} \times \vec{b}|$

Substituting the value of $|\vec{a} \times \vec{b}|$:

Area = $\dfrac{3\sqrt{3}}{2}$

Answer: $\dfrac{3\sqrt{3}}{2}$