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**Hint:**The center of mass is a position defined relative to an object or a system of objects. It is the average position of all parts of the given system, weighted according to their masses. There are situations in which the center of mass does not fall anywhere on the object.

**Complete step by step answer:**

Given, Mass of first particle is: ${m_1} = 10\,kg$

And the position vector is $\mathop {{r_1}}\limits^ \to = \mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge $

Mass of second particle is: ${m_2} = 30\,kg$

And the position vector is $\mathop {{r_2}}\limits^ \to = - \mathop i\limits^ \wedge - \mathop j\limits^ \wedge - \mathop k\limits^ \wedge $

$\text{Center of mass}= \dfrac{{{m_1}\mathop {{r_1}}\limits^ \to + {m_2}\mathop {{r_1}}\limits^ \to }}{{{m_1} + {m_2}}}$

$\Rightarrow \text{Center of mass} = \dfrac{{10\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right) + 30\left( { - \mathop i\limits^ \wedge - \mathop j\limits^ \wedge - \mathop k\limits^ \wedge } \right)}}{{10 + 30}}$

$\therefore \text{Center of mass} = - \left( {\dfrac{{\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge }}{2}} \right)$

**Therefore, option A is the correct answer.**

**Note:**The position vectors in physics are straight lines having one end fixed to a body and the other end attached to a moving point. They are used to describe the position of the point relative to the body. As the point moves, the position vector changes in length or in direction or in both length and direction. Note that the center of mass may sometimes lie outside the physical body. This happens in the case of hollow or open shaped objects like horseshoes.

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