Square of Difference of Two Vectors

Given,

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

Therefore,

$$|\vec a+\vec b|^2=21$$

$$|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b=21 \qquad (1)$$

Also,

$$|\vec a-\vec b|=3$$

Therefore,

$$|\vec a-\vec b|^2=9$$

$$|\vec a|^2+|\vec b|^2-2\vec a\cdot\vec b=9 \qquad (2)$$

Adding (1) and (2),

$$|\vec a|^2+|\vec b|^2=15 \qquad (3)$$

Subtracting (2) from (1),

$$4\vec a\cdot\vec b=12$$

$$\vec a\cdot\vec b=3 \qquad (4)$$

Since

$$\vec a\perp(\vec a-\vec b)$$

we have

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

$$|\vec a|^2-\vec a\cdot\vec b=0$$

Using (4),

$$|\vec a|^2=3$$

Hence from (3),

$$|\vec b|^2=12$$

Now,

$$\text{Area of } \triangle OPR=\frac12|\vec a\times(\vec a+\vec b)|$$

$$=\frac12|\vec a\times\vec b|$$

Also,

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

$$=(3)(12)-3^2$$

$$=36-9$$

$$=27$$

Therefore,

$$|\vec a\times\vec b|=3\sqrt3$$

Hence,

$$\text{Area of } \triangle OPR=\frac12(3\sqrt3)$$

$$=\boxed{\frac{3\sqrt3}{2}}$$