Let the three vectors be $\vec p=\overrightarrow{OP},\quad \vec q=\overrightarrow{OQ},\quad \vec r=\overrightarrow{OR}$ with $|\vec p|=|\vec q|=|\vec r|=A.$

From the figure, the three vectors are equally inclined, i.e. the angle between any two consecutive vectors is $120^\circ$.

Resultant vector: $\vec R=\vec p+\vec q+\vec r.$ Its magnitude is found from $|\vec R|^2=(\vec p+\vec q+\vec r)\cdot(\vec p+\vec q+\vec r).$

Expanding, $|\vec R|^2=|\vec p|^2+|\vec q|^2+|\vec r|^2+2\vec p\cdot\vec q+2\vec q\cdot\vec r+2\vec r\cdot\vec p.$

Since each magnitude is $A$ and angle between any pair is $120^\circ$, $\vec p\cdot\vec q=A^2\cos120^\circ=-\frac{A^2}{2}$ and similarly, $\vec q\cdot\vec r=-\frac{A^2}{2},\qquad \vec r\cdot\vec p=-\frac{A^2}{2}.$

Substitute: $|\vec R|^2=3A^2+2\left(-\frac{A^2}{2}-\frac{A^2}{2}-\frac{A^2}{2}\right)$ $=3A^2+2\left(-\frac{3A^2}{2}\right)$ $=3A^2-3A^2=0.$

Hence, $|\vec R|=0=A\sqrt{x}.$ Therefore, $\sqrt{x}=0\implies x=0.$

So the derived answer is $\boxed{0}.$

Comparison with stored correct answer:

• Stored correct answer: $3$
• Derived answer: $0$

These do not match. For three equal vectors separated by $120^\circ$, the resultant is zero, so the stored answer appears inconsistent with the stated geometry.